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The Chain Rule

State the chain rule for the composition of two functions.

Introduction

In addition to learning how to differentiate a variety of basic functions, we have also been developing our ability to use rules to differentiate certain algebraic combinations of them.

Example

State the rule(s) required to find the derivative of each of the following combinations of \(f(x) = \sin(x)\) and \(g(x) = x^2\): \[\begin{aligned}\end{aligned}\], \[\begin{aligned}\end{aligned}\] \[\begin{aligned}\end{aligned}\].

Solution

Finding \(s'\) uses the sum and constant multiple rules, because \(s(x) = 3g(x) - 5f(x)\). Determining \(p'\) requires the product rule, because \(p(x) = g(x) \cdot f(x)\). To calculate \(q'\) we use the quotient rule, because \(q(x) =\frac{f(x)}{g(x)}\).

There is one more natural way to combine basic functions algebraically, and that is by composing them. For instance, let's consider the function \[\begin{aligned}\end{aligned}\], and observe that any input \(x\) passes through a chain of functions. In the process that defines the function \(C(x)\), \(x\) is first squared, and then the sine of the result is taken. Using an arrow diagram, \[\begin{aligned}\end{aligned}\].

In terms of the elementary functions \(f\) and \(g\), we observe that \(x\) is the input for the function \(g\), and the result is used as the input for \(f\). We write \[\begin{aligned}\end{aligned}\] and say that \(C\) is the composition of \(f\) and \(g\). We will refer to \(g\), the function that is first applied to \(x\), as the inner function, while \(f\), the function that is applied to the result, is the outer function.

Given a composite function \(C(x) = f(g(x))\) that is built from differentiable functions \(f\) and \(g\), how do we compute \(C'(x)\) in terms of \(f\), \(g\), \(f'\), and \(g'\)? In the same way that the rate of change of a product of two functions, \(p(x) = f(x) \cdot g(x)\), depends on the behavior of both \(f\) and \(g\), it makes sense intuitively that the rate of change of a composite function \(C(x) = f(g(x))\) will also depend on some combination of \(f\) and \(g\) and their derivatives. The rule that describes how to compute \(C'\) in terms of \(f\) and \(g\) and their derivatives is called the chain rule.

Exploration

Condensed — the full section is in Boelkins, Active Calculus.

The chain rule

Often a composite function cannot be written in an alternate algebraic form. For instance, the function \(C(x) = \sin(x^2)\) cannot be expanded or otherwise rewritten, so it presents no alternate approaches to taking the derivative. But some composite functions can be expanded or simplified, and these provide a way to explore how the chain rule works.

Example

Let \(f(x) = -4x + 7\) and \(g(x) = 3x - 5\). Determine a formula for \(C(x) = f(g(x))\) and compute \(C'(x)\). How is \(C'\) related to \(f\) and \(g\) and their derivatives?

Solution

By the rules given for \(f\) and \(g\), \[\begin{aligned}C(x) =\mathstrut \amp f(g(x)) \\ =\mathstrut \amp f(3x-5) \\ =\mathstrut \amp -4(3x-5) + 7 \\ =\mathstrut \amp -12x + 20 + 7 \\ =\mathstrut \amp -12x + 27\end{aligned}\].

Thus, \(C'(x) = -12\). Noting that \(f'(x) = -4\) and \(g'(x) = 3\), we observe that \(C'\) appears to be the product of \(f'\) and \(g'\).

It may seem that Example is too elementary to illustrate how to differentiate a composite function. Linear functions are the simplest of all functions, and composing linear functions yields another linear function. While this example does not illustrate the full complexity of a composition of nonlinear functions, at the same time we remember that any differentiable function is locally linear, and thus any function with a derivative behaves like a line when viewed up close. The fact that the derivatives of the linear functions \(f\) and \(g\) are multiplied to find the derivative of their composition turns out to be a key insight.

We now consider a composition involving a nonlinear function.

Example

Let \(C(x) = \sin(2x)\). Use the double angle identity to rewrite \(C\) as a product of basic functions, and use the product rule to find \(C'\). Rewrite \(C'\) in the simplest form possible.

Solution

Using the double angle identity for the sine function, we write \[\begin{aligned}\end{aligned}\].

Applying the product rule and simplifying, we find \[\begin{aligned}\end{aligned}\].

Next, we recall that a double angle identity for the cosine tells us \[\begin{aligned}\end{aligned}\].

Substituting this result into our expression for \(C'(x)\), we now have that \[\begin{aligned}\end{aligned}\].

Condensed — the full section is in Boelkins, Active Calculus.

Using multiple rules simultaneously

The chain rule now joins the sum, constant multiple, product, and quotient rules in our collection of techniques for finding the derivative of a function through understanding its algebraic structure and the basic functions that constitute it. It takes practice to get comfortable applying multiple rules to differentiate a single function, but using proper notation and taking a few extra steps will help.

Example

Find a formula for the derivative of \(h(t) = 3^{t^2 + 2t}\sec^4(t)\).

Solution

We first observe that \(h\) is the product of two functions: \(h(t) = a(t) \cdot b(t)\), where \(a(t) = 3^{t^2 + 2t}\) and \(b(t) = \sec^4(t)\). We will need to use the product rule to differentiate \(h\). And because \(a\) and \(b\) are composite functions, we will need the chain rule. We therefore begin by computing \(a'(t)\) and \(b'(t)\).

Writing \(a(t) = f(g(t)) = 3^{t^2 + 2t}\), and finding the derivatives of \(f\) and \(g\), we have

\(f(t) = 3^t\)		\(g(t) = t^2 + 2t\)
\(f'(t) = 3^t \ln(3)\)		\(g'(t) = 2t+2\)
\(f'(g(t)) = 3^{t^2 + 2t}\ln(3)\)

Thus, by the chain rule, it follows that \(a'(t) = f'(g(t))g'(t) = 3^{t^2 + 2t}\ln(3) (2t+2)\).

Turning next to \(b\), we write \(b(t) = r(s(t)) = \sec^4(t)\) and find the derivatives of \(r\) and \(s\).

\(r(t) = t^4\)		\(s(t) = \sec(t)\)
\(r'(t) = 4t^3\)		\(s'(t) = \sec(t)\tan(t)\)
\(r'(s(t)) = 4\sec^3(t)\)

By the chain rule, \[\begin{aligned}\end{aligned}\].

Now we are finally ready to compute the derivative of the function \(h\). Recalling that \(h(t) = 3^{t^2 + 2t}\sec^4(t)\), by the product rule we have \[\begin{aligned}\end{aligned}\].

From our work above with \(a\) and \(b\), we know the derivatives of \(3^{t^2 + 2t}\) and \(\sec^4(t)\), and therefore \[\begin{aligned}\end{aligned}\].

The chain rule now adds substantially to our ability to compute derivatives. Whether we are finding the equation of the tangent line to a curve, the instantaneous velocity of a moving particle, or the instantaneous rate of change of a certain quantity, if the function under consideration is a composition, the chain rule is often an essential tool.

The composite version of basic function rules

As we gain more experience with differention, we will become more comfortable in simply writing down the derivative without taking multiple steps. This is particularly simple when the inner function is linear, since the derivative of a linear function is a constant.

Example

For each of the following composite functions whose inside function is linear, find the overall function's derivative using the chain rule: \(f(x) = (5x+7)^{10}\), \(g(x) = \tan(17x)\), and \(h(x) = e^{-3x}\).

Solution

For each of the three given functions, the derivative of the inner function is constant. By the chain rule, we see \[\begin{aligned}\end{aligned}\], \[\begin{aligned}\end{aligned}\] \[\begin{aligned}\end{aligned}\].

More generally, we can think about how each basic function rule has a corresponding chain rule version. The next example demonstrates this for two familiar functions.

Example

Develop a chain rule version of the two basic derivative rules that state \(\frac{d}{dx}[\sin(x)] = \cos(x)\) and \(\frac{d}{dx}[a^x] = a^x \ln(a)\).

Solution

To determine \[\begin{aligned}\end{aligned}\], where \(u\) is a differentiable function of \(x\), we use the chain rule with the sine function as the outer function. Applying the chain rule, we find that \[\begin{aligned}\end{aligned}\]. This rule is analogous to the basic derivative rule that \(\frac{d}{dx}[\sin(x)] = \cos(x)\).

Similarly, since \(\frac{d}{dx}[a^x] = a^x \ln(a)\), it follows by the chain rule that \[\begin{aligned}\end{aligned}\]. This rule is analogous to the basic derivative rule that \(\frac{d}{dx}[a^{x}] = a^{x} \ln(a)\).

An excellent exercise for getting comfortable with the derivative rules is to complete Example for every basic function. That is, write down a list of all the basic functions whose derivatives you know, and list their corresponding derivatives. Then, corresponding to each basic rule, write a composite function with the inner function being an unknown function \(u(x)\) and the outer function being a basic function. Finally, write the chain rule for the composite function, such as \(\frac{d}{dx}[\sin(u(x))] = \cos(u(x)) \cdot u'(x)\).

Summary

A composite function is one where the input variable \(x\) first passes through one function, and then the resulting output passes through another. For example, the function \(h(x) = 2^{\sin(x)}\) is composite since \(x \longrightarrow \sin(x) \longrightarrow 2^{\sin(x)}\).
Given a composite function \(C(x) = f(g(x))\) where \(f\) and \(g\) are differentiable functions, the chain rule tells us that \[\begin{aligned}\end{aligned}\].

Deriving the Chain Rule

When we have a function that is a composition of two or more functions, we could use all of the techniques we have already learned to differentiate it. However, using all of those techniques to break down a function into simpler parts that we are able to differentiate can get cumbersome. Instead, we use the chain rule, which states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.

To put this rule into context, let’s take a look at an example: \(h(x)=\text{sin}\ ({x}^{3}).\) We can think of the derivative of this function with respect to x as the rate of change of \(\text{sin}({x}^{3})\) relative to the change in \(x.\) Consequently, we want to know how \(\text{sin}({x}^{3})\) changes as \(x\) changes. We can think of this event as a chain reaction: As \(x\) changes, \({x}^{3}\) changes, which leads to a change in \(\text{sin}\ ({x}^{3}).\) This chain reaction gives us hints as to what is involved in computing the derivative of \(\ \text{sin}({x}^{3}).\) First of all, a change in \(x\) forcing a change in \({x}^{3}\) suggests that somehow the derivative of \({x}^{3}\) is involved. In addition, the change in \({x}^{3}\) forcing a change in \(\text{sin}({x}^{3})\) suggests that the derivative of \(\text{sin}(u)\) with respect to \(u,\) where \(u={x}^{3},\) is also part of the final derivative.

We can take a more formal look at the derivative of \(h(x)=\text{sin}\ ({x}^{3})\) by setting up the limit that would give us the derivative at a specific value \(a\) in the domain of \(h(x)=\text{sin}\ ({x}^{3}).\)

\[{h}^{'}(a)=\underset{x\to a}{\text{lim}}\frac{\text{sin}\ ({x}^{3})-\text{sin}({a}^{3})}{x-a}.\]

This expression does not seem particularly helpful; however, we can modify it by multiplying and dividing by the expression \({x}^{3}-{a}^{3}\) to obtain

\[{h}^{'}(a)=\underset{x\to a}{\text{lim}}\frac{\text{sin}\ ({x}^{3})-\text{sin}({a}^{3})}{{x}^{3}-{a}^{3}}\cdot \frac{{x}^{3}-{a}^{3}}{x-a}.\]

From the definition of the derivative, we can see that the second factor is the derivative of \({x}^{3}\) at \(x=a.\) That is,

\[\underset{x\to a}{\text{lim}}\frac{{x}^{3}-{a}^{3}}{x-a}=\frac{d}{dx}{({x}^{3})}_{x=a}=3{a}^{2}.\]

However, it might be a little more challenging to recognize that the first term is also a derivative. We can see this by letting \(u={x}^{3}\) and observing that as \(x\to a,u\to {a}^{3}\text{:}\)

Thus, \({h}^{'}(a)=\text{cos}({a}^{3})\cdot 3{a}^{2}.\)

Condensed — the full section is in OpenStax Calculus Volume 1.

The Chain and Power Rules Combined

We can now apply the chain rule to composite functions, but note that we often need to use it with other rules. For example, to find derivatives of functions of the form \(h(x)={(g(x))}^{n},\) we need to use the chain rule combined with the power rule. To do so, we can think of \(h(x)={(g(x))}^{n}\) as \(f(g(x))\) where \(f(x)={x}^{n}.\) Then \({f}^{'}(x)=n{x}^{n-1}.\) Thus, \({f}^{'}(g(x))=n{(g(x))}^{n-1}.\) This leads us to the derivative of a power function using the chain rule,

\[{h}^{'}(x)=n{(g(x))}^{n-1}{g}^{'}(x)\]

Example

Try it.

Find the derivative of \(h(x)=\frac{1}{{(3{x}^{2}+1)}^{2}}.\)

Solution

First, rewrite \(h(x)=\frac{1}{{(3{x}^{2}+1)}^{2}}={(3{x}^{2}+1)}^{-2}.\)

Applying the power rule with \(g(x)=3{x}^{2}+1,\) we have

\[{h}^{'}(x)=-2{(3{x}^{2}+1)}^{-3}(6x).\]

Rewriting back to the original form gives us

\[{h}^{'}(x)=\frac{-12x}{{(3{x}^{2}+1)}^{3}}.\]

Example

Try it.

Find the derivative of \(h(x)={\text{sin}}^{3}x.\)

Solution

First recall that \({\text{sin}}^{3}x={(\text{sin}\ x)}^{3},\) so we can rewrite \(h(x)={\text{sin}}^{3}x\) as \(h(x)={(\text{sin}\ x)}^{3}.\)

Applying the power rule with \(g(x)=\text{sin}\ x,\) we obtain

\[{h}^{'}(x)=3{(\text{sin}\ x)}^{2}\text{cos}\ x=3\ {\text{sin}}^{2}x\ \text{cos}\ x.\]

Example

Try it.

Find the equation of a line tangent to the graph of \(h(x)=\frac{1}{{(3x-5)}^{2}}\) at \(x=2.\)

Solution

Because we are finding an equation of a line, we need a point. The x-coordinate of the point is 2. To find the y-coordinate, substitute 2 into \(h(x).\) Since \(h(2)=\frac{1}{{(3(2)-5)}^{2}}=1,\) the point is \((2,1).\)

For the slope, we need \({h}^{'}(2).\) To find \({h}^{'}(x),\) first we rewrite \(h(x)={(3x-5)}^{-2}\) and apply the power rule to obtain

\[{h}^{'}(x)=-2{(3x-5)}^{-3}(3)=-6{(3x-5)}^{-3}.\]

By substituting, we have \({h}^{'}(2)=-6{(3(2)-5)}^{-3}=-6.\) Therefore, the line has equation \(y-1=-6(x-2).\) Rewriting, the equation of the line is \(y=-6x+13.\)

Combining the Chain Rule with Other Rules

Now that we can combine the chain rule and the power rule, we examine how to combine the chain rule with the other rules we have learned. In particular, we can use it with the formulas for the derivatives of trigonometric functions or with the product rule.

Example

Try it.

Find the derivative of \(h(x)=\text{cos}\ (g(x)).\)

Solution

Think of \(h(x)=\text{cos}(g(x))\) as \(f(g(x))\) where \(f(x)=\text{cos}\ x.\) Since \({f}^{'}(x)=\text{-}\text{sin}\ x.\) we have \({f}^{'}(g(x))=\text{-}\text{sin}\ (g(x)).\) Then we do the following calculation.

\[\begin{array}{lllll}{h}^{'}(x) & ={f}^{'}(g(x)){g}^{'}(x) & & & \text{Apply the chain rule.} \\ & =\text{-}\text{sin}\ (g(x)){g}^{'}(x) & & & \text{Substitute}\ {f}^{'}(g(x))=\text{-}\text{sin}\ (g(x)).\end{array}\]

Thus, the derivative of \(h(x)=\text{cos}\ (g(x))\) is given by \({h}^{'}(x)=\text{-}\text{sin}\ (g(x)){g}^{'}(x).\)

In the following example we apply the rule that we have just derived.

Example

Try it.

Find the derivative of \(h(x)=\text{cos}\ (5{x}^{2}).\)

Solution

Let \(g(x)=5{x}^{2}.\) Then \({g}^{'}(x)=10x.\) Using the result from the previous example,

\[\begin{array}{ll}{h}^{'}(x) & =\text{-}\text{sin}\ (5{x}^{2})\cdot 10x \\ & =-10x\ \text{sin}\ (5{x}^{2}).\end{array}\]

Example

Try it.

Find the derivative of \(h(x)=\text{sec}\ (4{x}^{5}+2x).\)

Solution

Apply the chain rule to \(h(x)=\text{sec}\ (g(x))\) to obtain

\[{h}^{'}(x)=\text{sec}(g(x))\ \text{tan}\ (g(x)){g}^{'}(x).\]

In this problem, \(g(x)=4{x}^{5}+2x,\) so we have \({g}^{'}(x)=20{x}^{4}+2.\) Therefore, we obtain

\[\begin{array}{ll}{h}^{'}(x) & =\text{sec}\ (4{x}^{5}+2x)\ \text{tan}\ (4{x}^{5}+2x)(20{x}^{4}+2) \\ & =(20{x}^{4}+2)\text{sec}\ (4{x}^{5}+2x)\ \text{tan}\ (4{x}^{5}+2x).\end{array}\]

At this point we provide a list of derivative formulas that may be obtained by applying the chain rule in conjunction with the formulas for derivatives of trigonometric functions. Their derivations are similar to those used in and . For convenience, formulas are also given in Leibniz’s notation, which some students find easier to remember. (We discuss the chain rule using Leibniz’s notation at the end of this section.) It is not absolutely necessary to memorize these as separate formulas as they are all applications of the chain rule to previously learned formulas.

Condensed — the full section is in OpenStax Calculus Volume 1.

Composites of Three or More Functions

We can now combine the chain rule with other rules for differentiating functions, but when we are differentiating the composition of three or more functions, we need to apply the chain rule more than once. If we look at this situation in general terms, we can generate a formula, but we do not need to remember it, as we can simply apply the chain rule multiple times.

In general terms, first we let

\[k(x)=h(f(g(x))).\]

Then, applying the chain rule once we obtain

\[{k}^{'}(x)=\frac{d}{dx}(h(f(g(x)))=h'(f(g(x)))\cdot \frac{d}{dx}f((g(x))).\]

Applying the chain rule again, we obtain

\[{k}^{'}(x)={h}^{'}(f(g(x)){f}^{'}(g(x)){g}^{'}(x)).\]

Notice that the derivative of the composition of three functions has three parts. (Similarly, the derivative of the composition of four functions has four parts, and so on.) Also, remember, we can always work from the outside in, taking one derivative at a time.

Example

Try it.

Find the derivative of \(k(x)={\text{cos}}^{4}({7x}^{2}+1).\)

Solution

First, rewrite \(k(x)\) as

\[k(x)={(\text{cos}\ (7{x}^{2}+1))}^{4}.\]

Then apply the power rule several times.

\[\begin{array}{lllll}{k}^{'}(x) & =4{(\text{cos}\ (7{x}^{2}+1))}^{3}(\frac{d}{dx}\text{cos}\ (7{x}^{2}+1)) & & & \text{Apply the chain rule.} \\ & =4{(\text{cos}\ (7{x}^{2}+1))}^{3}(\text{-}\text{sin}\ (7{x}^{2}+1))(\frac{d}{dx}(7{x}^{2}+1)) & & & \text{Apply the chain rule.} \\ & =4{(\text{cos}\ (7{x}^{2}+1))}^{3}(\text{-}\text{sin}\ (7{x}^{2}+1))(14x) & & & \text{Apply the chain rule.} \\ & =-56x\ \text{sin}\ (7{x}^{2}+1){\text{cos}}^{3}(7{x}^{2}+1) & & & \text{Simplify.}\end{array}\]

Example

Try it.

A particle moves along a coordinate axis. Its position at time t is given by \(s(t)=\text{sin}\ (2t)+\text{cos}\ (3t).\) What is the velocity of the particle at time \(t=\frac{\pi }{6}?\)

Solution

To find \(v(t),\) the velocity of the particle at time \(t,\) we must differentiate \(s(t).\) Thus,

\[v(t)={s}^{'}(t)=2\ \text{cos}\ (2t)-3\ \text{sin}\ (3t).\]

Substituting \(t=\frac{\pi }{6}\) into \(v(t),\) we obtain \(v(\frac{\pi }{6})=-2.\)

Condensed — the full section is in OpenStax Calculus Volume 1.

The Chain Rule Using Leibniz’s Notation

As with other derivatives that we have seen, we can express the chain rule using Leibniz’s notation. This notation for the chain rule is used heavily in physics applications.

\(\text{For}\ h(x)=f(g(x)),\) let \(u=g(x)\) and \(y=h(x)=f(u).\) Thus,

\[{h}^{'}(x)=\frac{dy}{dx},{f}^{'}(g(x))={f}^{'}(u)=\frac{dy}{du}\ \text{and}\ {g}^{'}(x)=\frac{du}{dx}.\]

Consequently,

\[\frac{dy}{dx}={h}^{'}(x)={f}^{'}(g(x)){g}^{'}(x)=\frac{dy}{du}\cdot \frac{du}{dx}.\]

Example

Try it.

Find the derivative of \(y={(\frac{x}{3x+2})}^{5}.\)

Solution

First, let \(u=\frac{x}{3x+2}.\) Thus, \(y={u}^{5}.\) Next, find \(\frac{du}{dx}\) and \(\frac{dy}{du}.\) Using the quotient rule,

\[\frac{du}{dx}=\frac{2}{{(3x+2)}^{2}}\]

and

\[\frac{dy}{du}=5{u}^{4}.\]

Finally, we put it all together.

\[\begin{array}{lllll}\frac{dy}{dx} & =\frac{dy}{du}\cdot \frac{du}{dx} & & & \text{Apply the chain rule.} \\ & =5{u}^{4}\cdot \frac{2}{{(3x+2)}^{2}} & & & \text{Substitute}\ \frac{dy}{du}=5{u}^{4}\ \text{and}\ \frac{du}{dx}=\frac{2}{{(3x+2)}^{2}}. \\ & =5{(\frac{x}{3x+2})}^{4}\cdot \frac{2}{{(3x+2)}^{2}} & & & \text{Substitute}\ u=\frac{x}{3x+2}. \\ & =\frac{10{x}^{4}}{{(3x+2)}^{6}} & & & \text{Simplify.}\end{array}\]

It is important to remember that, when using the Leibniz form of the chain rule, the final answer must be expressed entirely in terms of the original variable given in the problem.

Example

Try it.

Find the derivative of \(y=\text{tan}\ (4{x}^{2}-3x+1).\)

Solution

First, let \(u=4{x}^{2}-3x+1.\) Then \(y=\text{tan}\ u.\) Next, find \(\frac{du}{dx}\) and \(\frac{dy}{du}\text{:}\)

\[\frac{du}{dx}=8x-3\ \text{and}\ \frac{dy}{du}={\text{sec}}^{2}u.\]

Finally, we put it all together.

\[\begin{array}{lllll}\frac{dy}{dx} & =\frac{dy}{du}\cdot \frac{du}{dx} & & & \text{Apply the chain rule.} \\ & ={\text{sec}}^{2}u\cdot (8x-3) & & & \text{Use}\ \frac{du}{dx}=8x-3\ \text{and}\ \frac{dy}{du}={\text{sec}}^{2}u. \\ & ={\text{sec}}^{2}(4{x}^{2}-3x+1)\cdot (8x-3) & & & \text{Substitute}\ u=4{x}^{2}-3x+1.\end{array}\]

Key Concepts

The chain rule allows us to differentiate compositions of two or more functions. It states that for \(h(x)=f(g(x)),\)
\[{h}^{'}(x)={f}^{'}(g(x)){g}^{'}(x).\]
In Leibniz’s notation this rule takes the form
\[\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}.\]
We can use the chain rule with other rules that we have learned, and we can derive formulas for some of them.
The chain rule combines with the power rule to form a new rule:
\[\text{If}\ h(x)={(g(x))}^{n},\text{then}\ {h}^{'}(x)=n{(g(x))}^{n-1}{g}^{'}(x).\]
When applied to the composition of three functions, the chain rule can be expressed as follows: If \(h(x)=f(g(k(x))),\) then \({h}^{'}(x)={f}^{'}(g(k(x)){g}^{'}(k(x)){k}^{'}(x).\)

The Chain Rule

For the following exercises, given \(y=f(u)\) and \(u=g(x),\) find \(\frac{dy}{dx}\) by using Leibniz’s notation for the chain rule: \(\frac{dy}{dx}=\frac{dy}{du}\ \frac{du}{dx}.\)

For each of the following exercises,

decompose each function in the form \(y=f(u)\) and \(u=g(x),\) and
find \(\frac{dy}{dx}\) as a function of \(x.\)

For the following exercises, find \(\frac{dy}{dx}\) for each function.

For the following exercises, use the information in the following table to find \({h}^{'}(a)\) at the given value for \(a.\)

\(x\)	\(f(x)\)	\(f'(x)\)	\(g(x)\)	\(g'(x)\)
0	2	5	0	2
1	1	−2	3	0
2	4	4	1	−1
3	3	−3	2	3

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Each of the functions on the left below could be described as having a specific algebraic structure as noted on the right. Match each function with its corresponding structure. It is possible that not all structures are used; and it is possible that not all functions have a description for its algebraic structure.
If you do not find a match in the previous exercise, explain why.
You are riding a hot air balloon that is moving straight upward. You have access to an altimeter that tells you how high (in miles) above the ground the balloon is at a certain number of hours since the ride started [you can call it \(A(t)\)]. You also have access to a gauge that gives the air temperature (in degrees Fahrenheit) that you feel as a function of altitude [you can call it \(F(A)\)].
One hour into the ride, you happen to look at the screen and see the following: \(A'(1)=2\) and \(F'(2)=-16\). In your own words state:
1. What does \(A'(1) = 2\) tell you about your ride in the hot air balloon? Include units.
2. What does \(F'(2) = -16\) tell you about your ride in the hot air balloon? Include units.
3. Explain in your own words how the air temperature you feel is changing with time, one hour into the ride, and find that rate.
Consider the basic functions \(f(x) = x^3\) and \(g(x) = \sin(x)\).
1. Let \(h(x) = f(g(x))\). Find the exact instantaneous rate of change of \(h\) at the point where \(x = \frac{\pi}{4}\).
2. Which function is changing most rapidly at \(x = 0.25\): \(h(x) = f(g(x))\) or \(r(x) = g(f(x))\)? Why?
3. Let \(h(x) = f(g(x))\) and \(r(x) = g(f(x))\). Which of these functions has a derivative that is periodic? Why?
ਜਵਾਬ ਦਿਓ
1. Since \(h(x) = f(g(x))\), by the chain rule we know that \(h'(x) = f'(g(x)) \cdot g'(x)\). Using the facts that \(f(x)=x^3\) and \(g(x) = \sin(x)\), it follows \(f'(x) = 3x^2\) and \(f'(g(x)) = 3\sin^2(x)\), and \(g'(x) = \cos(x)\). Therefore, \[\begin{aligned}\end{aligned}\] and \(h'\left( \frac{\pi}{4} \right) = 3 \left( \frac{1}{\sqrt{2}} \right)^2 \cdot \frac{1}{\sqrt{2}} = \frac{3}{2\sqrt{2}}\).
2. Note that \(r'(x) = g'(f(x)) \cdot f'(x) = \cos(x^3) \cdot 3x^2\). Thus, we have \(r'(0.25) = \cos(0.25^3) \cdot 3(0.25)^2 \approx 0.1875\), while using our work from (a), \(h'(0.25) = 3\sin^2(0.25) \cdot \cos(0.25) \approx 0.1779\), so since \(|h'(0.25) \lt r'(0.25)|\), we see that \(r\) is changing more rapidly.
3. We have observed in (a) and (b) that \(h'(x) = 3\sin^2(x) \cdot \cos(x)\) and \(r'(x) = \cos(x^3) \cdot 3x^2\). Because both parts of the product in \(h'(x)\) are periodic functions, \(h'(x)\) is periodic (with period \(2\pi\)). However, \(r'(x)\) is not periodic, since \(3x^2\) grows without bound as \(x\) increases.
Let \(u(x)\) be a differentiable function. For each of the following functions, determine the derivative. Each response will involve \(u\) and/or \(u'\).
1. \(p(x) = e^{u(x)}\)
2. \(q(x) = u(e^x)\)
3. \(r(x) = \cot(u(x))\)
4. \(s(x) = u(\cot(x))\)
5. \(a(x) = u(x^4)\)
6. \(b(x) = (u(x))^4\)
ਜਵਾਬ ਦਿਓ
1. By the chain rule, \(p'(x) = e^{u(x)} \cdot u'(x)\).
2. By the chain rule, \(q'(x) = u'(e^x) \cdot e^x\).
3. By the chain rule, \(r'(x) = -\csc^2(u(x) \cdot u'(x)\).
4. By the chain rule, \(s'(x) = u'(\cot(x)) \cdot (-\csc^2(x))\).
5. By the chain rule, \(a'(x) = u'(x^4) \cdot 4x^3\).
6. By the chain rule, \(b'(x) = 4(u(x))^3 \cdot u'(x)\).
Let functions \(p\) and \(q\) be the piecewise linear functions given by their respective graphs in Figure. Use the graphs to answer the following questions.
1. Let \(C(x) = p(q(x))\). Determine \(C'(0)\) and \(C'(3)\).
2. Let \(Y(x) = q(q(x))\) and \(Z(x) = q(p(x))\). Determine \(Y'(-2)\) and \(Z'(0)\).
ਜਵਾਬ ਦਿਓ
1. By the chain rule, \(C'(x) = p'(q(x)) \cdot q'(x)\), so \(C'(0) = p'(q(0)) \cdot q'(0)\) and \(C'(3) = p'(q(3)) \cdot q'(3)\). From the graph, we see that \(q(0) = 2\), and thus \(C'(0) = p'(2) \cdot q'(0)\). Reading the slopes of the lines from the graph, it follows \(C'(0) = 2 \cdot 0 = 0\). Similarly, \(q(3) = 0\), so \(C'(3) = p'(0) \cdot q'(3) = \frac{1}{2} \cdot (-1) = -\frac{1}{2}\).
2. Since \(Y(x) = q(q(x))\), the chain rule implies that \(Y'(x) = q'(q(x)) \cdot q'(x)\), and thus \(Y'(-2) = q'(q(-2)) \cdot q'(-2) = q'(-1) \cdot q'(-2)\). But \(q'(-1)\) does not exist, so \(Y'(-2)\) also fails to exist. Using \(Z(x) = q(p(x))\) and the chain rule, we have \(Z'(x) = q'(p(x)) \cdot p'(x)\). Therefore \(Z'(0) = q'(p(0)) \cdot p'(0) = q'(-0.5) \cdot p'(0) = 0 \cdot 0.5 = 0\).
If a spherical tank of radius 4 feet has \(h\) feet of water present in the tank, then the volume of water in the tank is given by the formula \[\begin{aligned}\end{aligned}\].
1. At what instantaneous rate is the volume of water in the tank changing with respect to the height of the water at the instant \(h = 1\)? What are the units on this quantity?
2. Now suppose that the height of water in the tank is being regulated by an inflow and outflow (e.g., a faucet and a drain) so that the height of the water at time \(t\) is given by the rule \(h(t) = \sin(\pi t) + 1\), where \(t\) is measured in hours (and \(h\) is still measured in feet). At what rate is the height of the water changing with respect to time at the instant \(t = 2\)?
3. Continuing under the assumptions in (b), at what instantaneous rate is the volume of water in the tank changing with respect to time at the instant \(t = 2\)?
4. What are the main differences between the rates found in (a) and (c)? Include a discussion of the relevant units.
ਜਵਾਬ ਦਿਓ
1. We are asked to find \(\frac{dV}{dh}\) (the derivative of \(V\) with respect to the independent variable \(h\)) at the instant \(h = 1\). If we first expand \(V\) to write \[\begin{aligned}\end{aligned}\] we find that \[\begin{aligned}\end{aligned}\], with units cubic feet per foot. Evaluating at \(h=1\), we get \[\begin{aligned}\end{aligned}\]
2. Since \(h(t) = \sin(\pi t) + 1\), the instantaneous rate of change of \(h\) with respect to \(t\) is \(h'(t) = \pi \cos(\pi t)\), and therefore \(h'(2) = \pi \cos(2\pi) = \pi\) feet per hour.
3. First we want to find \(\frac{dV}{dt}\), the derivative of \(V\) with respect to the independent variable \(t\). Here we can write \(V\) as a function of \(t\) by writing \(V(t) = \frac{\pi}{3}\left(12(h(t))^2 - (h(t))^3\right)\), where \(h(t) = \sin(\pi t) + 1\). By the chain rule, it follows that \[\begin{aligned}\end{aligned}\] and hence using \(h(t) = \sin(\pi t) + 1\) and \(h'(t) = \pi \cos(\pi t)\), \[\begin{aligned}\end{aligned}\] with units feet per hour.
  Now we want to find \(\frac{dV}{dt}\) when \(t=2\). Using our most recent work, we have \[\begin{aligned}\end{aligned}\] cubic feet per hour.
4. In (a) we are determining the instantaneous rate at which the volume changes as we increase the height of the water in the tank, while in (c) we are finding the instantaneous rate at which volume changes as we increase time. The first calculation is a direct one, while the second is more indirect, as it involves the composition of \(V\) with \(h\) as a function of time. In the latter case, both \(h\) and \(V\) are functions of time, which allows us to think not only of how \(V\) is changing as a function of \(h\), but also as a function of \(t\).
Consider the basic functions \(f(x) = x^3\) and \(g(x) = \sin(x)\).
1. Let \(h(x) = f(g(x))\). Find the exact instantaneous rate of change of \(h\) at the point where \(x = \frac{\pi}{4}\).
2. Which function is changing most rapidly at \(x = 0.25\): \(h(x) = f(g(x))\) or \(r(x) = g(f(x))\)? Why?
3. Let \(h(x) = f(g(x))\) and \(r(x) = g(f(x))\). Which of these functions has a derivative that is periodic? Why?
ਜਵਾਬ ਦਿਓ
1. Since \(h(x) = f(g(x))\), by the chain rule we know that \(h'(x) = f'(g(x)) \cdot g'(x)\). Using the facts that \(f(x)=x^3\) and \(g(x) = \sin(x)\), it follows \(f'(x) = 3x^2\) and \(f'(g(x)) = 3\sin^2(x)\), and \(g'(x) = \cos(x)\). Therefore, \[\begin{aligned}\end{aligned}\] and \(h'\left( \frac{\pi}{4} \right) = 3 \left( \frac{1}{\sqrt{2}} \right)^2 \cdot \frac{1}{\sqrt{2}} = \frac{3}{2\sqrt{2}}\).
2. Note that \(r'(x) = g'(f(x)) \cdot f'(x) = \cos(x^3) \cdot 3x^2\). Thus, we have \(r'(0.25) = \cos(0.25^3) \cdot 3(0.25)^2 \approx 0.1875\), while using our work from (a), \(h'(0.25) = 3\sin^2(0.25) \cdot \cos(0.25) \approx 0.1779\), so since \(|h'(0.25) \lt r'(0.25)|\), we see that \(r\) is changing more rapidly.
3. We have observed in (a) and (b) that \(h'(x) = 3\sin^2(x) \cdot \cos(x)\) and \(r'(x) = \cos(x^3) \cdot 3x^2\). Because both parts of the product in \(h'(x)\) are periodic functions, \(h'(x)\) is periodic (with period \(2\pi\)). However, \(r'(x)\) is not periodic, since \(3x^2\) grows without bound as \(x\) increases.
Let \(u(x)\) be a differentiable function. For each of the following functions, determine the derivative. Each response will involve \(u\) and/or \(u'\).
1. \(p(x) = e^{u(x)}\)
2. \(q(x) = u(e^x)\)
3. \(r(x) = \cot(u(x))\)
4. \(s(x) = u(\cot(x))\)
5. \(a(x) = u(x^4)\)
6. \(b(x) = u^4(x)\)
ਜਵਾਬ ਦਿਓ
1. By the chain rule, \(p'(x) = e^{u(x)} \cdot u'(x)\).
2. By the chain rule, \(q'(x) = u'(e^x) \cdot e^x\).
3. By the chain rule, \(r'(x) = -\csc^2(u(x) \cdot u'(x)\).
4. By the chain rule, \(s'(x) = u'(\cot(x)) \cdot (-\csc^2(x))\).
5. By the chain rule, \(a'(x) = u'(x^4) \cdot 4x^3\).
6. By the chain rule, \(b'(x) = 4(u(x))^3 \cdot u'(x)\).
Let functions \(p\) and \(q\) be the piecewise linear functions given by their respective graphs in Figure. Use the graphs to answer the following questions.
1. Let \(C(x) = p(q(x))\). Determine \(C'(0)\) and \(C'(3)\).
2. Let \(Y(x) = q(q(x))\) and \(Z(x) = q(p(x))\). Determine \(Y'(-2)\) and \(Z'(0)\).
ਜਵਾਬ ਦਿਓ
1. By the chain rule, \(C'(x) = p'(q(x)) \cdot q'(x)\), so \(C'(0) = p'(q(0)) \cdot q'(0)\) and \(C'(3) = p'(q(3)) \cdot q'(3)\). From the graph, we see that \(q(0) = 2\), and thus \(C'(0) = p'(2) \cdot q'(0)\). Reading the slopes of the lines from the graph, it follows \(C'(0) = 2 \cdot 0 = 0\). Similarly, \(q(3) = 0\), so \(C'(3) = p'(0) \cdot q'(3) = \frac{1}{2} \cdot (-1) = -\frac{1}{2}\).
2. Since \(Y(x) = q(q(x))\), the chain rule implies that \(Y'(x) = q'(q(x)) \cdot q'(x)\), and thus \(Y'(-2) = q'(q(-2)) \cdot q'(-2) = q'(-1) \cdot q'(-2)\). But \(q'(-1)\) does not exist, so \(Y'(-2)\) also fails to exist. Using \(Z(x) = q(p(x))\) and the chain rule, we have \(Z'(x) = q'(p(x)) \cdot p'(x)\). Therefore \(Z'(0) = q'(p(0)) \cdot p'(0) = q'(-0.5) \cdot p'(0) = 0 \cdot 0.5 = 0\).
If a spherical tank of radius 4 feet has \(h\) feet of water present in the tank, then the volume of water in the tank is given by the formula \[\begin{aligned}\end{aligned}\].
1. At what instantaneous rate is the volume of water in the tank changing with respect to the height of the water at the instant \(h = 1\)? What are the units on this quantity?
2. Now suppose that the height of water in the tank is being regulated by an inflow and outflow (e.g., a faucet and a drain) so that the height of the water at time \(t\) is given by the rule \(h(t) = \sin(\pi t) + 1\), where \(t\) is measured in hours (and \(h\) is still measured in feet). At what rate is the height of the water changing with respect to time at the instant \(t = 2\)?
3. Continuing under the assumptions in (b), at what instantaneous rate is the volume of water in the tank changing with respect to time at the instant \(t = 2\)?
4. What are the main differences between the rates found in (a) and (c)? Include a discussion of the relevant units.
ਜਵਾਬ ਦਿਓ
1. We are asked to find \(\frac{dV}{dh}\) (the derivative of \(V\) with respect to the independent variable \(h\)) at the instant \(h = 1\). If we first expand \(V\) to write \[\begin{aligned}\end{aligned}\] we find that \[\begin{aligned}\end{aligned}\], with units cubic feet per foot. Evaluating at \(h=1\), we get \[\begin{aligned}\end{aligned}\]
2. Since \(h(t) = \sin(\pi t) + 1\), the instantaneous rate of change of \(h\) with respect to \(t\) is \(h'(t) = \pi \cos(\pi t)\), and therefore \(h'(2) = \pi \cos(2\pi) = \pi\) feet per hour.
3. First we want to find \(\frac{dV}{dt}\), the derivative of \(V\) with respect to the independent variable \(t\). Here we can write \(V\) as a function of \(t\) by writing \(V(t) = \frac{\pi}{3}\left(12(h(t))^2 - (h(t))^3\right)\), where \(h(t) = \sin(\pi t) + 1\). By the chain rule, it follows that \[\begin{aligned}\end{aligned}\] and hence using \(h(t) = \sin(\pi t) + 1\) and \(h'(t) = \pi \cos(\pi t)\), \[\begin{aligned}\end{aligned}\] with units feet per hour.
  Now we want to find \(\frac{dV}{dt}\) when \(t=2\). Using our most recent work, we have \[\begin{aligned}\end{aligned}\] cubic feet per hour.
4. In (a) we are determining the instantaneous rate at which the volume changes as we increase the height of the water in the tank, while in (c) we are finding the instantaneous rate at which volume changes as we increase time. The first calculation is a direct one, while the second is more indirect, as it involves the composition of \(V\) with \(h\) as a function of time. In the latter case, both \(h\) and \(V\) are functions of time, which allows us to think not only of how \(V\) is changing as a function of \(h\), but also as a function of \(t\).
Find the derivative of \(h(x)=\frac{1}{{(3{x}^{2}+1)}^{2}}.\)

ਜਵਾਬ ਦਿਓ

First, rewrite \(h(x)=\frac{1}{{(3{x}^{2}+1)}^{2}}={(3{x}^{2}+1)}^{-2}.\)
Applying the power rule with \(g(x)=3{x}^{2}+1,\) we have
\[{h}^{'}(x)=-2{(3{x}^{2}+1)}^{-3}(6x).\]
Rewriting back to the original form gives us
\[{h}^{'}(x)=\frac{-12x}{{(3{x}^{2}+1)}^{3}}.\]
Find the derivative of \(h(x)={(2{x}^{3}+2x-1)}^{4}.\)

ਜਵਾਬ ਦਿਓ

\({h}^{'}(x)=4{(2{x}^{3}+2x-1)}^{3}(6{x}^{2}+2)=8(3{x}^{2}+1){(2{x}^{3}+2x-1)}^{3}\)
Find the derivative of \(h(x)={\text{sin}}^{3}x.\)

ਜਵਾਬ ਦਿਓ

First recall that \({\text{sin}}^{3}x={(\text{sin}\ x)}^{3},\) so we can rewrite \(h(x)={\text{sin}}^{3}x\) as \(h(x)={(\text{sin}\ x)}^{3}.\)
Applying the power rule with \(g(x)=\text{sin}\ x,\) we obtain
\[{h}^{'}(x)=3{(\text{sin}\ x)}^{2}\text{cos}\ x=3\ {\text{sin}}^{2}x\ \text{cos}\ x.\]
Find the equation of a line tangent to the graph of \(h(x)=\frac{1}{{(3x-5)}^{2}}\) at \(x=2.\)

ਜਵਾਬ ਦਿਓ

Because we are finding an equation of a line, we need a point. The x-coordinate of the point is 2. To find the y-coordinate, substitute 2 into \(h(x).\) Since \(h(2)=\frac{1}{{(3(2)-5)}^{2}}=1,\) the point is \((2,1).\)
For the slope, we need \({h}^{'}(2).\) To find \({h}^{'}(x),\) first we rewrite \(h(x)={(3x-5)}^{-2}\) and apply the power rule to obtain
\[{h}^{'}(x)=-2{(3x-5)}^{-3}(3)=-6{(3x-5)}^{-3}.\]
By substituting, we have \({h}^{'}(2)=-6{(3(2)-5)}^{-3}=-6.\) Therefore, the line has equation \(y-1=-6(x-2).\) Rewriting, the equation of the line is \(y=-6x+13.\)
Find an equation of the line tangent to the graph of \(f(x)={({x}^{2}-2)}^{3}\) at \(x=-2.\)

ਜਵਾਬ ਦਿਓ

\(y=-48x-88\)
Find the derivative of \(h(x)=\text{cos}\ (g(x)).\)

ਜਵਾਬ ਦਿਓ

Think of \(h(x)=\text{cos}(g(x))\) as \(f(g(x))\) where \(f(x)=\text{cos}\ x.\) Since \({f}^{'}(x)=\text{-}\text{sin}\ x.\) we have \({f}^{'}(g(x))=\text{-}\text{sin}\ (g(x)).\) Then we do the following calculation.
\[\begin{array}{lllll}{h}^{'}(x) & ={f}^{'}(g(x)){g}^{'}(x) & & & \text{Apply the chain rule.} \\ & =\text{-}\text{sin}\ (g(x)){g}^{'}(x) & & & \text{Substitute}\ {f}^{'}(g(x))=\text{-}\text{sin}\ (g(x)).\end{array}\]
Thus, the derivative of \(h(x)=\text{cos}\ (g(x))\) is given by \({h}^{'}(x)=\text{-}\text{sin}\ (g(x)){g}^{'}(x).\)
Find the derivative of \(h(x)=\text{cos}\ (5{x}^{2}).\)

ਜਵਾਬ ਦਿਓ

Let \(g(x)=5{x}^{2}.\) Then \({g}^{'}(x)=10x.\) Using the result from the previous example,
\[\begin{array}{ll}{h}^{'}(x) & =\text{-}\text{sin}\ (5{x}^{2})\cdot 10x \\ & =-10x\ \text{sin}\ (5{x}^{2}).\end{array}\]
Find the derivative of \(h(x)=\text{sec}\ (4{x}^{5}+2x).\)

ਜਵਾਬ ਦਿਓ

Apply the chain rule to \(h(x)=\text{sec}\ (g(x))\) to obtain
\[{h}^{'}(x)=\text{sec}(g(x))\ \text{tan}\ (g(x)){g}^{'}(x).\]
In this problem, \(g(x)=4{x}^{5}+2x,\) so we have \({g}^{'}(x)=20{x}^{4}+2.\) Therefore, we obtain
\[\begin{array}{ll}{h}^{'}(x) & =\text{sec}\ (4{x}^{5}+2x)\ \text{tan}\ (4{x}^{5}+2x)(20{x}^{4}+2) \\ & =(20{x}^{4}+2)\text{sec}\ (4{x}^{5}+2x)\ \text{tan}\ (4{x}^{5}+2x).\end{array}\]
Find the derivative of \(h(x)=\text{sin}(7x+2).\)

ਜਵਾਬ ਦਿਓ

\({h}^{'}(x)=7\ \text{cos}\ (7x+2)\)
Find the derivative of \(h(x)={(2x+1)}^{5}{(3x-2)}^{7}.\)

ਜਵਾਬ ਦਿਓ

First apply the product rule, then apply the chain rule to each term of the product.
\[\begin{array}{lllll}{h}^{'}(x) & =\frac{d}{dx}({(2x+1)}^{5})\cdot {(3x-2)}^{7}+\frac{d}{dx}({(3x-2)}^{7})\cdot {(2x+1)}^{5} & & & \text{Apply the product rule.} \\ & =5{(2x+1)}^{4}\cdot 2\cdot {(3x-2)}^{7}+7{(3x-2)}^{6}\cdot 3\cdot {(2x+1)}^{5} & & & \text{Apply the chain rule.} \\ & =10{(2x+1)}^{4}{(3x-2)}^{7}+21{(3x-2)}^{6}{(2x+1)}^{5} & & & \text{Simplify.} \\ & ={(2x+1)}^{4}{(3x-2)}^{6}(10(3x-2)+21(2x+1)) & & & \text{Factor out}\ {(2x+1)}^{4}{(3x-2)}^{6}. \\ & ={(2x+1)}^{4}{(3x-2)}^{6}(72x+1) & & & \text{Simplify.}\end{array}\]
Find the derivative of \(h(x)=\frac{x}{{(2x+3)}^{3}}.\)

ਜਵਾਬ ਦਿਓ

\({h}^{'}(x)=\frac{3-4x}{{(2x+3)}^{4}}\)
Find the derivative of \(k(x)={\text{cos}}^{4}({7x}^{2}+1).\)

ਜਵਾਬ ਦਿਓ

First, rewrite \(k(x)\) as
\[k(x)={(\text{cos}\ (7{x}^{2}+1))}^{4}.\]
Then apply the power rule several times.
\[\begin{array}{lllll}{k}^{'}(x) & =4{(\text{cos}\ (7{x}^{2}+1))}^{3}(\frac{d}{dx}\text{cos}\ (7{x}^{2}+1)) & & & \text{Apply the chain rule.} \\ & =4{(\text{cos}\ (7{x}^{2}+1))}^{3}(\text{-}\text{sin}\ (7{x}^{2}+1))(\frac{d}{dx}(7{x}^{2}+1)) & & & \text{Apply the chain rule.} \\ & =4{(\text{cos}\ (7{x}^{2}+1))}^{3}(\text{-}\text{sin}\ (7{x}^{2}+1))(14x) & & & \text{Apply the chain rule.} \\ & =-56x\ \text{sin}\ (7{x}^{2}+1){\text{cos}}^{3}(7{x}^{2}+1) & & & \text{Simplify.}\end{array}\]
Find the derivative of \(h(x)={\text{sin}}^{6}({x}^{3}).\)

ਜਵਾਬ ਦਿਓ

\({h}^{'}(x)=18{x}^{2}{\text{sin}}^{5}({x}^{3})\ \text{cos}\ ({x}^{3})\)
A particle moves along a coordinate axis. Its position at time t is given by \(s(t)=\text{sin}\ (2t)+\text{cos}\ (3t).\) What is the velocity of the particle at time \(t=\frac{\pi }{6}?\)

ਜਵਾਬ ਦਿਓ

To find \(v(t),\) the velocity of the particle at time \(t,\) we must differentiate \(s(t).\) Thus,
\[v(t)={s}^{'}(t)=2\ \text{cos}\ (2t)-3\ \text{sin}\ (3t).\]
Substituting \(t=\frac{\pi }{6}\) into \(v(t),\) we obtain \(v(\frac{\pi }{6})=-2.\)
A particle moves along a coordinate axis. Its position at time \(t\) is given by \(s(t)=\text{sin}(4t).\) Find its acceleration at time \(t.\)

ਜਵਾਬ ਦਿਓ

\(a(t)=-16\ \text{sin}(4t)\)
Let \(h(x)=f(g(x)).\) If \(g(1)=4,{g}^{'}(1)=3,\) and \({f}^{'}(4)=7,\) find \({h}^{'}(1).\)

ਜਵਾਬ ਦਿਓ

Use the chain rule, then substitute.
\[\begin{array}{lllll}{h}^{'}(1) & ={f}^{'}(g(1)){g}^{'}(1) & & & \text{Apply the chain rule.} \\ & ={f}^{'}(4)\cdot 3 & & & \text{Substitute}\ g(1)=4\ \text{and}\ {g}^{'}(1)=3. \\ & =7\cdot 3 & & & \text{Substitute}\ f'(4)=7. \\ & =21 & & & \text{Simplify.}\end{array}\]
Given \(h(x)=f(g(x)).\) If \(g(2)=-3,{g}^{'}(2)=4,\) and \({f}^{'}(-3)=7,\) find \({h}^{'}(2).\)

ਜਵਾਬ ਦਿਓ

\(28\)
Find the derivative of \(y={(\frac{x}{3x+2})}^{5}.\)

ਜਵਾਬ ਦਿਓ

First, let \(u=\frac{x}{3x+2}.\) Thus, \(y={u}^{5}.\) Next, find \(\frac{du}{dx}\) and \(\frac{dy}{du}.\) Using the quotient rule,
\[\frac{du}{dx}=\frac{2}{{(3x+2)}^{2}}\]
and
\[\frac{dy}{du}=5{u}^{4}.\]
Finally, we put it all together.
\[\begin{array}{lllll}\frac{dy}{dx} & =\frac{dy}{du}\cdot \frac{du}{dx} & & & \text{Apply the chain rule.} \\ & =5{u}^{4}\cdot \frac{2}{{(3x+2)}^{2}} & & & \text{Substitute}\ \frac{dy}{du}=5{u}^{4}\ \text{and}\ \frac{du}{dx}=\frac{2}{{(3x+2)}^{2}}. \\ & =5{(\frac{x}{3x+2})}^{4}\cdot \frac{2}{{(3x+2)}^{2}} & & & \text{Substitute}\ u=\frac{x}{3x+2}. \\ & =\frac{10{x}^{4}}{{(3x+2)}^{6}} & & & \text{Simplify.}\end{array}\]
It is important to remember that, when using the Leibniz form of the chain rule, the final answer must be expressed entirely in terms of the original variable given in the problem.
Find the derivative of \(y=\text{tan}\ (4{x}^{2}-3x+1).\)

ਜਵਾਬ ਦਿਓ

First, let \(u=4{x}^{2}-3x+1.\) Then \(y=\text{tan}\ u.\) Next, find \(\frac{du}{dx}\) and \(\frac{dy}{du}\text{:}\)
\[\frac{du}{dx}=8x-3\ \text{and}\ \frac{dy}{du}={\text{sec}}^{2}u.\]
Finally, we put it all together.
\[\begin{array}{lllll}\frac{dy}{dx} & =\frac{dy}{du}\cdot \frac{du}{dx} & & & \text{Apply the chain rule.} \\ & ={\text{sec}}^{2}u\cdot (8x-3) & & & \text{Use}\ \frac{du}{dx}=8x-3\ \text{and}\ \frac{dy}{du}={\text{sec}}^{2}u. \\ & ={\text{sec}}^{2}(4{x}^{2}-3x+1)\cdot (8x-3) & & & \text{Substitute}\ u=4{x}^{2}-3x+1.\end{array}\]
Use Leibniz’s notation to find the derivative of \(y=\text{cos}\ ({x}^{3}).\) Make sure that the final answer is expressed entirely in terms of the variable \(x.\)

ਜਵਾਬ ਦਿਓ

\(\frac{dy}{dx}=-3{x}^{2}\text{sin}\ ({x}^{3})\)
\(y=3u-6,u=2{x}^{2}\)
\(y=6{u}^{3},u=7x-4\)

ਜਵਾਬ ਦਿਓ

\(18{u}^{2}\cdot 7=18{(7x-4)}^{2}\cdot 7\)
\(y=\text{sin}\ u,u=5x-1\)
\(y=\text{cos}\ u,u=\frac{\text{-}x}{8}\)

ਜਵਾਬ ਦਿਓ

\(\text{-}\text{sin}\ u\cdot \frac{-1}{8}=\text{-}\text{sin}\ (\frac{\text{-}x}{8})\cdot \frac{-1}{8}\)
\(y=\text{tan}\ u,u=9x+2\)
\(y=\sqrt{4u+3},u={x}^{2}-6x\)

ਜਵਾਬ ਦਿਓ

\(\frac{8x-24}{2\sqrt{4u+3}}=\frac{4x-12}{\sqrt{4{x}^{2}-24x+3}}\)
\(y={(3x-2)}^{6}\)
\(y={(3{x}^{2}+1)}^{3}\)

ਜਵਾਬ ਦਿਓ

a. \(u=3{x}^{2}+1;\) b. \(18x{(3{x}^{2}+1)}^{2}\)
\(y={\text{sin}}^{5}(x)\)

Symbols used here

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

Euler's number
2.71828…, the base whose exponential is its own derivative.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: The Chain Rule

State the chain rule for the composition of two functions.
Apply the chain rule together with the power rule.
Apply the chain rule and the product/quotient rules correctly in combination when both are necessary.
Recognize the chain rule for a composition of three or more functions.
Describe the proof of the chain rule.
To differentiate
Find
Find

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

ਆਪਣਾ ਹੀ ਕੋਸ਼ਿਸ਼ ਕਰੋ

Parts of this page are adapted from Boelkins, Active Calculus (CC BY-SA 4.0), OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

ਹੋਰ ਵਿੱਚ Calculus

Limits Derivatives Integrals Definite integrals Taylor series Series and sums Maxima and minima The chain rule Implicit differentiation Related rates and optimisation Integration techniques: substitution, parts, partial fractions Applications of integration: area, volume, arc length Infinite series and convergence tests