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Elementary derivative rules

In Chapter, we developed the concept of the derivative of a function. We now know that the derivative f' of a function f measures the instantaneous rate of change of f with respect to x.

Introduction

In Chapter, we developed the concept of the derivative of a function. We now know that the derivative \(f'\) of a function \(f\) measures the instantaneous rate of change of \(f\) with respect to \(x\). The derivative also tells us the slope of the tangent line to \(y=f(x)\) at any given value of \(x\). So far, we have focused on interpreting the derivative graphically or, in the context of a physical setting, as a meaningful rate of change. To calculate the value of the derivative at a specific point, we have relied on the limit definition of the derivative, \[\begin{aligned}\end{aligned}\].

In this chapter, we investigate how the limit definition of the derivative leads to interesting patterns and rules that enable us to find a formula for \(f'(x)\) quickly, without using the limit definition directly. For example, we would like to apply shortcuts to differentiate a function such as \(g(x) = 4x^7 - \sin(x) + 3e^x\) simply by observation.

Exploration

Some Key Notation

In addition to our usual \(f'\) notation, there are other ways to denote the derivative of a function, as well as the instruction to take the derivative. If we are thinking about the relationship between \(y\) and \(x\), we sometimes denote the derivative of \(y\) with respect to \(x\) by the symbol \[\begin{aligned}\end{aligned}\] which we read dee-y dee-x. For example, if \(y = x^2\), we'll write that the derivative is \(\frac{dy}{dx} = 2x\). This notation comes from the fact that the derivative is related to the slope of a line, and slope is measured by \(\frac{\Delta y}{\Delta x}\). Note that while we read \(\frac{\Delta y}{\Delta x}\) as change in \(y\) over change in \(x\), we view \(\frac{dy}{dx}\) as a single symbol, not a quotient of two quantities.

We use a variant of this notation as the instruction to take the derivative. In particular, \[\begin{aligned}\end{aligned}\] means take the derivative of the quantity in \(\Box\) with respect to \(x\). For example, we may write \(\frac{d}{dx}[x^2] = 2x\).

It is important to note that the independent variable can be different from \(x\). If we have \(f(z) = z^2\), we then write \(f'(z) = 2z\). Similarly, if \(y = t^2\), we say \(\frac{dy}{dt} = 2t\). And it is also true that \(\frac{d}{dq}[q^2] = 2q\). This notation may also be used for second derivatives: \(f''(z) = \frac{d}{dz}\left[\frac{df}{dz}\right] = \frac{d^2 f}{dz^2}\).

In what follows, we'll build a repertoire of functions for which we can quickly compute the derivative.

Constant, Power, and Exponential Functions

So far, we know the derivative formula for two important classes of functions: constant functions and power functions. If \(f(x) = c\) is a constant function, its graph is a horizontal line with slope zero at every point. Thus, \(\frac{d}{dx}[c] = 0\). We summarize this with the following rule.

For any real number \(c\), if \(f(x) = c\), then \(f'(x) = 0\).

Example

If \(f(x) = 7\), then \(f'(x) = 0\). Similarly, \(\frac{d}{dx} [\sqrt{3}] = 0\).

In your work in Preview Activity, you conjectured that for any positive integer \(n\), if \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\). This rule can be formally proved for any positive integer \(n\), and even for any nonzero real number (positive or negative).

For any nonzero real number \(n\), if \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\).

Example

Using the rule for power functions, we can compute the following derivatives. If \(g(z) = z^{-3}\), then \(g'(z) = -3z^{-4}\). Similarly, if \(h(t) = t^{7/5}\), then \(\frac{dh}{dt} = \frac{7}{5}t^{2/5}\), and \(\frac{d}{dq} [q^{\pi}] = \pi q^{\pi - 1}\).

It will be helpful to have a derivative formula for one more type of basic function. For now, we simply state this rule without explanation or justification; we explore why this rule is true in one of the exercises and we will encounter graphical reasoning for why the rule is plausible in Preview Activity.

For any positive real number \(a\), if \(f(x) = a^x\), then \(f'(x) = a^x \ln(a)\).

Example

If \(f(x) = 2^x\), then \(f'(x) = 2^x \ln(2)\). Similarly, for \(p(t) = 10^t\), \(p'(t) = 10^t \ln(10)\). It is especially important to note that when \(a = e\), where \(e\) is the base of the natural logarithm function, we have that \[\begin{aligned}\end{aligned}\] since \(\ln(e) = 1\). This is an extremely important property of the function \(e^x\): its derivative function is itself!

Note carefully the distinction between power functions and exponential functions: in power functions, the variable is in the base, as in \(x^2\), while in exponential functions, the variable is in the power, as in \(2^x\). As we can see from the rules, this makes a big difference in the form of the derivative.

Constant Multiples and Sums of Functions

Next we will learn how to compute the derivative of a function constructed as an algebraic combination of basic functions. For instance, we'd like to be able to take the derivative of a polynomial function such as \[\begin{aligned}\end{aligned}\], which is a sum of constant multiples of powers of \(t\). To that end, we develop two new rules: the Constant Multiple Rule and the Sum Rule.

How is the derivative of \(y = kf(x)\) related to the derivative of \(y = f(x)\)? Recall that when we multiply a function by a constant \(k\), we vertically stretch the graph by a factor of \(|k|\) (and reflect the graph across \(y = 0\) if \(k \lt 0\)). This vertical stretch affects the slope of the graph, so the slope of the function \(y = kf(x)\) is \(k\) times as steep as the slope of \(y = f(x)\). Thus, when we multiply a function by a factor of \(k\), we change the value of its derivative by a factor of \(k\) as well. The Constant Multiple Rule can be formally proved as a consequence of properties of limits, using the limit definition of the derivative.

For any real number \(k\), if \(f(x)\) is a differentiable function with derivative \(f'(x)\), then \(\frac{d}{dx}[k f(x)] = k f'(x)\).

In words, this rule says that the derivative of a constant times a function is the constant times the derivative of the function.

Example

If \(g(t) = 3 \cdot 5^t\), we have \(g'(t) = 3 \cdot 5^t \ln(5)\). Similarly, \(\frac{d}{dz} [5z^{-2}] = 5 (-2z^{-3})\).

Next we examine a sum of two functions. If we have \(y = f(x)\) and \(y = g(x)\), we can compute a new function \(y = (f+g)(x)\) by adding the outputs of the two functions: \((f+g)(x) = f(x) + g(x)\). Not only is the value of the new function the sum of the values of the two known functions, but the slope of the new function is the sum of the slopes of the known functions. Therefore Like the Constant Multiple Rule, the Sum Rule can be formally proved as a consequence of properties of limits, using the limit definition of the derivative. , we arrive at the following Sum Rule for derivatives:

If \(f(x)\) and \(g(x)\) are differentiable functions with derivatives \(f'(x)\) and \(g'(x)\) respectively, then \(\frac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x)\).

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

Summary

Given a differentiable function \(y = f(x)\), we can express the derivative of \(f\) in several different notations: \(f'(x)\), \(\frac{df}{dx}\), \(\frac{dy}{dx}\), and \(\frac{d}{dx}[f(x)]\).
The limit definition of the derivative leads to patterns among certain families of functions that enable us to compute derivative formulas without resorting directly to the limit definition. For example, if \(f\) is a power function of the form \(f(x) = x^n\), then \(f'(x) = nx^{n-1}\) for any real number \(n\) other than 0.
We have stated a rule for derivatives of exponential functions in the same spirit as the rule for power functions: for any positive real number \(a\), if \(f(x) = a^x\), then \(f'(x) = a^x \ln(a)\).
If we are given a constant multiple of a function whose derivative we know, or a sum of functions whose derivatives we know, the Constant Multiple and Sum Rules make it straightforward to compute the derivative of the overall function. More formally, if \(f(x)\) and \(g(x)\) are differentiable with derivatives \(f'(x)\) and \(g'(x)\) and \(a\) and \(b\) are constants, then \[\begin{aligned}\end{aligned}\].

Practice (11)

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

Consider the function \(f(x) = 2x + 1\), and observe that this function is linear. What is the value of the slope of \(f(x) = 2x + 1\) at any given point on the graph? What should be the formula for \(f'(x)\)?
For the linear function \(g(x)=x\), what is the value of the slope of \(g\) at any point on the graph? What should be the formula for \(g'(x)\)?
If \(h(x)=7-\frac{2}{3}x\), what do you expect will be the formula for \(h'(x)\)?
Suppose that \(p(x) = 3x - 5\) and \(q(x) = 4x + 8\). If \(r(x) = p(x) + q(x)\), what do you think will be the formula for \(r'(x)\)? Why?
Suppose that \(p(x) = 3x - 5\). If \(s(x) = -4 \cdot p(x)\), what do you think will be the formula for \(s'(x)\)? Why?
Let \(f\) and \(g\) be differentiable functions for which the following information is known: \(f(2) = 5\), \(g(2) = -3\), \(f'(2) = -1/2\), \(g'(2) = 2\).
1. Let \(h\) be the new function defined by the rule \(h(x) = 3f(x) - 4g(x)\). Determine \(h(2)\) and \(h'(2)\).
2. Find an equation for the tangent line to \(y = h(x)\) at the point \((2,h(2))\).
3. Let \(p\) be the function defined by the rule \(p(x) = -2f(x) + \frac{1}{2}g(x)\). Is \(p\) increasing, decreasing, or neither at \(a = 2\)? Why?
4. Estimate the value of \(p(2.03)\) by using the local linearization of \(p\) at the point \((2,p(2))\).
Revelar la respuesta
1. First, since \(h(x) = 3f(x) - 4g(x)\), we know \(h(2) = 3f(2) - 4g(2) = 3(5) - 4(-3) = 27\). Next, by the sum and constant multiple rules, \(h'(x) = 3f'(x) - 4g'(x)\). Hence, using the given derivative values, \(h'(2) = 3f'(2) - 4g'(2) = 3(-1/2) - 4(2) = -19/2\).
2. The tangent line at the point \((2,h(2))\) has equation \(L(x) = h(2) + h'(2)(x-2)\). Using the information determined in (a), we see \(L(x) = 27 - \frac{19}{2}(x-2)\).
3. Since \(p(x) = -2f(x) + \frac{1}{2}g(x)\), it follows that \(p'(x) = -2f'(x) + \frac{1}{2}g'(x)\) by the sum and constant multiple rules. Using the given function and derivative values for \(f\) and \(g\) at \(x=2\), we see that \(p(2) = -2f(2) + \frac{1}{2}g(2) = -2(5) + \frac{1}{2}(-3) = -\frac{23}{2}\) and \(p'(2) = -2(-\frac{1}{2}) + \frac{1}{2}(2)) = 2\). Since \(p'(2) = 2 \gt 0\), we conclude that \(p\) is increasing at \(x=2\).
4. First, the linearization of \(p\) at \(a = 2\) is \(L(x) = p(2) + p'(2)(x-2) = -\frac{23}{2} + 2(x-2)\). Using the linearization to estimate \(p(2.03)\), we find that \[\begin{aligned}\end{aligned}\].
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. At what values of \(x\) is \(p\) not differentiable? At what values of \(x\) is \(q\) not differentiable? Why?
2. Let \(r(x) = p(x) + 2q(x)\). At what values of \(x\) is \(r\) not differentiable? Why?
3. Determine \(r'(-2)\) and \(r'(0)\).
4. Find an equation for the tangent line to \(y = r(x)\) at the point \((2,r(2))\).
Revelar la respuesta
1. The function \(p\) is differentiable at points where its graph is locally linear or smooth. Since there are sharp corners in the graph of \(p\) at \(x=-1\) and \(x=1\), we see that \(p\) is not differentiable at those points. Similarly, the sharp corners in \(q\) at \(x=-1\) and \(x=1\) mean that \(q\) is not differentiable at these points, either.
2. In order for \(r\) to be differentiable at a point, both \(p\) and \(q\) must be differentiable at the point. So \(r\) is not differentiable at \(x=-1\) and \(x=1\).
3. Using the sum and scalar multiple rules, we have \(r'(x) = p'(x) + 2q'(x)\). The slope of \(p\) at \(x=-2\) is \(-2\) and the slope of \(q\) at \(x=-2\) is \(3\), so \(p'(-2) = -2\) and \(q'(-2) = 3\). It follows that \('(-2) = p'(-2) + 2q'(-2) = (-2) + 2(3) = 4\). Similarly, the slope of \(p\) at \(x=0\) is \(\frac{1}{2}\) and the slope of \(q\) at \(x=0\) is \(0\), so \(p'(0) = \frac{1}{2}\) and \(q'(0) = 0\). It follows that \(r'(0) = p'(0) + 2q'(0) = \frac{1}{2} + 2(0) = 1\).
4. Note that \(r(2) = p(2)+2q(2) = 2+2(1) = 4\) and \(r'(2) = p'(2)+2q'(2) = 2+2(-1) = 0\). Thus, the line tangent to the graph of \(r\) at \(x=2\) is \(y = r(2) + r'(2)(x-2) = 4\).
Consider the functions \(r(t) = t^t\) and \(s(t) = \arccos(t)\), for which you are given the facts that \(r'(t) = t^t(\ln(t) + 1)\) and \(s'(t) = -\frac{1}{\sqrt{1-t^2}}\). Do not be concerned with where these derivative formulas come from. We restrict our interest in both functions to the domain \(0 \lt t \lt 1\).
1. Let \(w(t) = 3t^t - 2\arccos(t)\). Determine \(w'(t)\).
2. Find an equation for the tangent line to \(y = w(t)\) at the point \((\frac{1}{2}, w(\frac{1}{2}))\).
3. Let \(v(t) = t^t + \arccos(t)\). Is \(v\) increasing or decreasing at the instant \(t = \frac{1}{2}\)? Why?
Revelar la respuesta
1. We note that \(w\) has the structure \(w(t) = 3r(t) - 2s(t)\), so by the sum and constant multiple rules, \(w'(t) = 3r'(t) - 2s'(t)\). Applying the given information about \(r'\) and \(s'\), \(w'(t) = 3t^t(\ln(t) + 1) + 2\frac{1}{\sqrt{1-t^2}}\).
2. We see that \(w(\frac{1}{2}) = 3(\frac{1}{2})^{\frac{1}{2}} - 2\arccos(\frac{1}{2}) = 3 \cdot \frac{1}{\sqrt{2}} - 2 \cdot \frac{\pi}{3} = \frac{3}{\sqrt{2}} - \frac{2\pi}{3}\). Similarly, \[\begin{aligned}w'\left(\frac{1}{2}\right) &= 3\left(\frac{1}{2}\right)^{\frac{1}{2}}\left(\ln\left(\frac{1}{2}\right) + 1\right) + 2\frac{1}{\sqrt{1-(\frac{1}{2})^2}} \\ &= \frac{3}{\sqrt{2}}\left(\ln\left(\frac{1}{2}\right) + 1\right) + \frac{4}{\sqrt{3}}\end{aligned}\]. Using these values, we find that the tangent line to \(w\) at the point \((\frac{1}{2},w(\frac{1}{2})\) is \[\begin{aligned}\end{aligned}\].
3. To determine whether \(v\) is increasing or decreasing at \(t = \frac{1}{2}\), we compute \(v'(\frac{1}{2})\). First, \(v'(t) = t^t(\ln(t) + 1) - \frac{1}{\sqrt{1-t^2}}\). Hence, \[\begin{aligned}\end{aligned}\]. Since \(v'(\frac{1}{2})\) is negative, \(v\) is decreasing at \(t = \frac{1}{2}\).
Let \(f\) and \(g\) be differentiable functions for which the following information is known: \(f(2) = 5\), \(g(2) = -3\), \(f'(2) = -1/2\), \(g'(2) = 2\).
1. Let \(h\) be the new function defined by the rule \(h(x) = 3f(x) - 4g(x)\). Determine \(h(2)\) and \(h'(2)\).
2. Find an equation for the tangent line to \(y = h(x)\) at the point \((2,h(2))\).
3. Let \(p\) be the function defined by the rule \(p(x) = -2f(x) + \frac{1}{2}g(x)\). Is \(p\) increasing, decreasing, or neither at \(a = 2\)? Why?
4. Estimate the value of \(p(2.03)\) by using the local linearization of \(p\) at the point \((2,p(2))\).
Revelar la respuesta
1. First, since \(h(x) = 3f(x) - 4g(x)\), we know \(h(2) = 3f(2) - 4g(2) = 3(5) - 4(-3) = 27\). Next, by the sum and constant multiple rules, \(h'(x) = 3f'(x) - 4g'(x)\). Hence, using the given derivative values, \(h'(2) = 3f'(2) - 4g'(2) = 3(-1/2) - 4(2) = -19/2\).
2. The tangent line at the point \((2,h(2))\) has equation \(L(x) = h(2) + h'(2)(x-2)\). Using the information determined in (a), we see \(L(x) = 27 - \frac{19}{2}(x-2)\).
3. Since \(p(x) = -2f(x) + \frac{1}{2}g(x)\), it follows that \(p'(x) = -2f'(x) + \frac{1}{2}g'(x)\) by the sum and constant multiple rules. Using the given function and derivative values for \(f\) and \(g\) at \(x=2\), we see that \(p(2) = -2f(2) + \frac{1}{2}g(2) = -2(5) + \frac{1}{2}(-3) = -\frac{23}{2}\) and \(p'(2) = -2(-\frac{1}{2}) + \frac{1}{2}(2)) = 2\). Since \(p'(2) = 2 \gt 0\), we conclude that \(p\) is increasing at \(x=2\).
4. First, the linearization of \(p\) at \(a = 2\) is \(L(x) = p(2) + p'(2)(x-2) = -\frac{23}{2} + 2(x-2)\). Using the linearization to estimate \(p(2.03)\), we find that \[\begin{aligned}\end{aligned}\].
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. At what values of \(x\) is \(p\) not differentiable? At what values of \(x\) is \(q\) not differentiable? Why?
2. Let \(r(x) = p(x) + 2q(x)\). At what values of \(x\) is \(r\) not differentiable? Why?
3. Determine \(r'(-2)\) and \(r'(0)\).
4. Find an equation for the tangent line to \(y = r(x)\) at the point \((2,r(2))\).
Revelar la respuesta
1. The function \(p\) is differentiable at points where its graph is locally linear or smooth. Since there are sharp corners in the graph of \(p\) at \(x=-1\) and \(x=1\), we see that \(p\) is not differentiable at those points. Similarly, the sharp corners in \(q\) at \(x=-1\) and \(x=1\) mean that \(q\) is not differentiable at these points, either.
2. In order for \(r\) to be differentiable at a point, both \(p\) and \(q\) must be differentiable at the point. So \(r\) is not differentiable at \(x=-1\) and \(x=1\).
3. Using the sum and scalar multiple rules, we have \(r'(x) = p'(x) + 2q'(x)\). The slope of \(p\) at \(x=-2\) is \(-2\) and the slope of \(q\) at \(x=-2\) is \(3\), so \(p'(-2) = -2\) and \(q'(-2) = 3\). It follows that \('(-2) = p'(-2) + 2q'(-2) = (-2) + 2(3) = 4\). Similarly, the slope of \(p\) at \(x=0\) is \(\frac{1}{2}\) and the slope of \(q\) at \(x=0\) is \(0\), so \(p'(0) = \frac{1}{2}\) and \(q'(0) = 0\). It follows that \(r'(0) = p'(0) + 2q'(0) = \frac{1}{2} + 2(0) = 1\).
4. Note that \(r(2) = p(2)+2q(2) = 2+2(1) = 4\) and \(r'(2) = p'(2)+2q'(2) = 2+2(-1) = 0\). Thus, the line tangent to the graph of \(r\) at \(x=2\) is \(y = r(2) + r'(2)(x-2) = 4\).
Consider the functions \(r(t) = t^t\) and \(s(t) = \arccos(t)\), for which you are given the facts that \(r'(t) = t^t(\ln(t) + 1)\) and \(s'(t) = -\frac{1}{\sqrt{1-t^2}}\). Do not be concerned with where these derivative formulas come from. We restrict our interest in both functions to the domain \(0 \lt t \lt 1\).
1. Let \(w(t) = 3t^t - 2\arccos(t)\). Determine \(w'(t)\).
2. Find an equation for the tangent line to \(y = w(t)\) at the point \((\frac{1}{2}, w(\frac{1}{2}))\).
3. Let \(v(t) = t^t + \arccos(t)\). Is \(v\) increasing or decreasing at the instant \(t = \frac{1}{2}\)? Why?
Revelar la respuesta
1. We note that \(w\) has the structure \(w(t) = 3r(t) - 2s(t)\), so by the sum and constant multiple rules, \(w'(t) = 3r'(t) - 2s'(t)\). Applying the given information about \(r'\) and \(s'\), \(w'(t) = 3t^t(\ln(t) + 1) + 2\frac{1}{\sqrt{1-t^2}}\).
2. We see that \(w(\frac{1}{2}) = 3(\frac{1}{2})^{\frac{1}{2}} - 2\arccos(\frac{1}{2}) = 3 \cdot \frac{1}{\sqrt{2}} - 2 \cdot \frac{\pi}{3} = \frac{3}{\sqrt{2}} - \frac{2\pi}{3}\). Similarly, \[\begin{aligned}w'\left(\frac{1}{2}\right) &= 3\left(\frac{1}{2}\right)^{\frac{1}{2}}\left(\ln\left(\frac{1}{2}\right) + 1\right) + 2\frac{1}{\sqrt{1-(\frac{1}{2})^2}} \\ &= \frac{3}{\sqrt{2}}\left(\ln\left(\frac{1}{2}\right) + 1\right) + \frac{4}{\sqrt{3}}\end{aligned}\]. Using these values, we find that the tangent line to \(w\) at the point \((\frac{1}{2},w(\frac{1}{2})\) is \[\begin{aligned}\end{aligned}\].
3. To determine whether \(v\) is increasing or decreasing at \(t = \frac{1}{2}\), we compute \(v'(\frac{1}{2})\). First, \(v'(t) = t^t(\ln(t) + 1) - \frac{1}{\sqrt{1-t^2}}\). Hence, \[\begin{aligned}\end{aligned}\]. Since \(v'(\frac{1}{2})\) is negative, \(v\) is decreasing at \(t = \frac{1}{2}\).

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\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.

\arcsin,\ \sin^{-1}

inverse sine
The angle whose sine is the given value (and likewise arccos, arctan).

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.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\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: Elementary derivative rules

What are alternate notations for the derivative?
How can we use the algebraic structure of a function f(x) to compute a formula for f'(x)?
What is the derivative of a power function of the form f(x) = x^n? What is the derivative of an exponential function of form f(x) = a^x?
If we know the derivative of y = f(x), what is the derivative of y = k f(x), where k is a constant?
If we know the derivatives of y = f(x) and y = g(x), how do we compute the derivative of y = f(x) + g(x)?

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.

Prueba tu propio

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

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