maths.free › Calculus › 3. Derivatives › The Derivative as a Function

The Derivative as a Function

Define the derivative function of a given function.

Derivative Functions

The derivative function gives the derivative of a function at each point in the domain of the original function for which the derivative is defined. We can formally define a derivative function as follows.

A function \(f(x)\) is said to be differentiable at \(a\) if \({f}^{'}(a)\) exists. More generally, a function is said to be differentiable on \(S\) if it is differentiable at every point in an open set \(S,\) and a differentiable function is one in which \({f}^{'}(x)\) exists on its domain.

In the next few examples we use to find the derivative of a function.

Example

Try it.

Find the derivative of \(f(x)=\sqrt{x}.\)

Solution

Start directly with the definition of the derivative function. Use .

\[\begin{array}{lllll}{f}^{'}(x) & =\underset{h\to 0}{\text{lim}}\frac{\sqrt{x+h}-\sqrt{x}}{h} & & & \begin{array}{l}\text{Substitute}\ f(x+h)=\sqrt{x+h}\ \text{and}\ f(x)=\sqrt{x} \\ \text{into}\ {f}^{'}(x)=\underset{h\to 0}{\text{lim}}\frac{f(x+h)-f(x)}{h}.\end{array} \\ & =\underset{h\to 0}{\text{lim}}\frac{\sqrt{x+h}-\sqrt{x}}{h}\cdot \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} & & & \begin{array}{l}\text{Multiply numerator and denominator by} \\ \sqrt{x+h}+\sqrt{x}\ \text{without distributing in the} \\ \text{denominator.}\end{array} \\ & =\underset{h\to 0}{\text{lim}}\frac{h}{h(\sqrt{x+h}+\sqrt{x})} & & & \text{Multiply the numerators and simplify.} \\ & =\underset{h\to 0}{\text{lim}}\frac{1}{(\sqrt{x+h}+\sqrt{x})} & & & \text{Cancel the}\ h. \\ & =\frac{1}{2\sqrt{x}} & & & \text{Evaluate the limit.}\end{array}\]

We use a variety of different notations to express the derivative of a function. In we showed that if \(f(x)={x}^{2}-2x,\) then \({f}^{'}(x)=2x-2.\) If we had expressed this function in the form \(y={x}^{2}-2x,\) we could have expressed the derivative as \({y}^{'}=2x-2\) or \(\frac{dy}{dx}=2x-2.\) We could have conveyed the same information by writing \(\frac{d}{dx}({x}^{2}-2x)=2x-2.\) Thus, for the function \(y=f(x),\) each of the following notations represents the derivative of \(f(x)\text{:}\)

\[{f}^{'}(x),\ \frac{dy}{dx},\ {y}^{'},\ \frac{d}{dx}(f(x)).\]\[\frac{dy}{dx}=\underset{\text{\Delta }x\to 0}{\text{lim}}\frac{\text{\Delta }y}{\text{\Delta }x}.\]

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

Graphing a Derivative

We have already discussed how to graph a function, so given the equation of a function or the equation of a derivative function, we could graph it. Given both, we would expect to see a correspondence between the graphs of these two functions, since \({f}^{'}(x)\) gives the rate of change of a function \(f(x)\) (or slope of the tangent line to \(f(x)).\)

In we found that for \(f(x)=\sqrt{x},{f}^{'}(x)=\frac{1}{2\sqrt{x}}.\) If we graph these functions on the same axes, as in , we can use the graphs to understand the relationship between these two functions. First, we notice that \(f(x)\) is increasing over its entire domain, which means that the slopes of its tangent lines at all points are positive. Consequently, we expect \({f}^{'}(x)>0\) for all values of \(x\) in its domain. Furthermore, as \(x\) increases, the slopes of the tangent lines to \(f(x)\) are decreasing and we expect to see a corresponding decrease in \({f}^{'}(x).\) We also observe that \(f'(0)\) is undefined and that \(\underset{x\to {0}^{+}}{\text{lim}}{f}^{'}(x)=\text{+}\infty ,\) corresponding to a vertical tangent to \(f(x)\) at \(0.\)

In we found that for \(f(x)={x}^{2}-2x,{f}^{'}(x)=2x-2.\) The graphs of these functions are shown in . Observe that \(f(x)\) is decreasing for \(x<1.\) For these same values of \(x,{f}^{'}(x)<0.\) For values of \(x>1,f(x)\) is increasing and \({f}^{'}(x)>0.\) Also, \(f(x)\) has a horizontal tangent at \(x=1\) and \({f}^{'}(1)=0.\)

Example

Try it.

Use the following graph of \(f(x)\) to sketch a graph of \({f}^{'}(x).\)

Solution

The solution is shown in the following graph. Observe that \(f(x)\) is increasing and \({f}^{'}(x)>0\) on \((-2,3).\) Also, \(f(x)\) is decreasing and \({f}^{'}(x)<0\) on \((\text{-}\infty ,-2)\) and on \((3,\text{+}\infty ).\) Also note that \(f(x)\) has horizontal tangents at \(-2\) and \(3,\) and \({f}^{'}(-2)=0\) and \({f}^{'}(3)=0.\)

Derivatives and Continuity

Now that we can graph a derivative, let’s examine the behavior of the graphs. First, we consider the relationship between differentiability and continuity. We will see that if a function is differentiable at a point, it must be continuous there; however, a function that is continuous at a point need not be differentiable at that point. In fact, a function may be continuous at a point and fail to be differentiable at the point for one of several reasons.

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

Higher-Order Derivatives

The derivative of a function is itself a function, so we can find the derivative of a derivative. For example, the derivative of a position function is the rate of change of position, or velocity. The derivative of velocity is the rate of change of velocity, which is acceleration. The new function obtained by differentiating the derivative is called the second derivative. Furthermore, we can continue to take derivatives to obtain the third derivative, fourth derivative, and so on. Collectively, these are referred to as higher-order derivatives. The notation for the higher-order derivatives of \(y=f(x)\) can be expressed in any of the following forms:

\[{f}^{″}(x),\ f\text{‴}(x),{f}^{(4)}(x)\text{,\ldots },{f}^{(n)}(x)\]\[{y}^{″}(x),y\text{‴}(x),{y}^{(4)}(x)\text{,\ldots },{y}^{(n)}(x)\]\[\frac{{d}^{2}y}{d{x}^{2}},\frac{{d}^{3}y}{d{x}^{3}},\frac{{d}^{4}y}{d{x}^{4}}\text{,\ldots },\frac{{d}^{n}y}{d{x}^{n}}.\]

It is interesting to note that the notation for \(\frac{{d}^{2}y}{d{x}^{2}}\) may be viewed as an attempt to express \(\frac{d}{dx}(\frac{dy}{dx})\) more compactly. Analogously, \(\frac{d}{dx}(\frac{d}{dx}(\frac{dy}{dx}))=\frac{d}{dx}(\frac{{d}^{2}y}{d{x}^{2}})=\frac{{d}^{3}y}{d{x}^{3}}.\)

Example

Try it.

For \(f(x)=2{x}^{2}-3x+1,\) find \({f}^{″}(x).\)

Solution

First find \({f}^{'}(x).\)

\[\begin{array}{lllll}{f}^{'}(x) & =\underset{h\to 0}{\text{lim}}\frac{(2{(x+h)}^{2}-3(x+h)+1)-(2{x}^{2}-3x+1)}{h} & & & \begin{array}{l}\text{Substitute}\ f(x)=2{x}^{2}-3x+1 \\ \text{and} \\ f(x+h)=2{(x+h)}^{2}-3(x+h)+1 \\ \text{into}\ {f}^{'}(x)=\underset{h\to 0}{\text{lim}}\frac{f(x+h)-f(x)}{h}.\end{array} \\ & =\underset{h\to 0}{\text{lim}}\frac{4xh+2{h}^{2}-3h}{h} & & & \text{Simplify the numerator.} \\ & =\underset{h\to 0}{\text{lim}}(4x+2h-3) & & & \begin{array}{l}\text{Factor out the}\ h\ \text{in the numerator} \\ \text{and cancel with the}\ h\ \text{in the} \\ \text{denominator.}\end{array} \\ & =4x-3 & & & \text{Take the limit.}\end{array}\]

Next, find \({f}^{″}(x)\) by taking the derivative of \({f}^{'}(x)=4x-3.\)

\[\begin{array}{lllll}{f}^{″}(x) & =\underset{h\to 0}{\text{lim}}\frac{{f}^{'}(x+h)-{f}^{'}(x)}{h} & & & \begin{array}{l}\text{Use}\ {f}^{'}(x)=\underset{h\to 0}{\text{lim}}\frac{f(x+h)-f(x)}{h}\ \text{with}\ {f}^{'}(x)\ \text{in} \\ \text{place of}\ f(x).\end{array} \\ & =\underset{h\to 0}{\text{lim}}\frac{(4(x+h)-3)-(4x-3)}{h} & & & \begin{array}{l}\text{Substitute}\ {f}^{'}(x+h)=4(x+h)-3\ \text{and} \\ {f}^{'}(x)=4x-3.\end{array} \\ & =\underset{h\to 0}{\text{lim}}4 & & & \text{Simplify.} \\ & =4 & & & \text{Take the limit.}\end{array}\]

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

Key Concepts

The derivative of a function \(f(x)\) is the function whose value at \(x\) is \({f}^{'}(x).\)
The graph of a derivative of a function \(f(x)\) is related to the graph of \(f(x).\) Where \(f(x)\) has a tangent line with positive slope, \({f}^{'}(x)>0.\) Where \(f(x)\) has a tangent line with negative slope, \({f}^{'}(x)<0.\) Where \(f(x)\) has a horizontal tangent line, \({f}^{'}(x)=0.\)
If a function is differentiable at a point, then it is continuous at that point. A function is not differentiable at a point if it is not continuous at the point, if it has a vertical tangent line at the point, or if the graph has a sharp corner or cusp.
Higher-order derivatives are derivatives of derivatives, from the second derivative to the \(n\text{th}\) derivative.

The Derivative as a Function

For the following exercises, use the definition of a derivative to find \({f}^{'}(x).\)

For the following exercises, use the graph of \(y=f(x)\) to sketch the graph of its derivative \({f}^{'}(x).\)

For the following exercises, the given limit represents the derivative of a function \(y=f(x)\) at \(x=a.\) Find \(f(x)\) and \(a.\)

For the following functions,

sketch the graph and
use the definition of a derivative to show that the function is not differentiable at \(x=1.\)

For the following graphs,

determine for which values of \(x=a\) the \(\underset{x\to a}{\text{lim}}f(x)\) exists but \(f\) is not continuous at \(x=a,\) and
determine for which values of \(x=a\) the function is continuous but not differentiable at \(x=a.\)

For the following functions, use \({f}^{″}(x)=\underset{h\to 0}{\text{lim}}\frac{{f}^{'}(x+h)-{f}^{'}(x)}{h}\) to find \({f}^{″}(x).\)

For the following exercises, use a calculator to graph \(f(x).\) Determine the function \({f}^{'}(x),\) then use a calculator to graph \({f}^{'}(x).\)

\(\frac{f(x+h)-f(x)}{h}\)
\({f}^{'}(x)=\underset{h\to 0}{\text{lim}}\frac{f(x+h)-f(x)}{h}\)

Time (seconds)	Height (meters)
\(0\)	\(0\)
\(1\)	\(2\)
\(2\)	\(4\)
\(3\)	\(13\)
\(4\)	\(25\)
\(5\)	\(32\)

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

Practice (40)

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

Find the derivative of \(f(x)=\sqrt{x}.\)

Giải đáp

Start directly with the definition of the derivative function. Use .
\[\begin{array}{lllll}{f}^{'}(x) & =\underset{h\to 0}{\text{lim}}\frac{\sqrt{x+h}-\sqrt{x}}{h} & & & \begin{array}{l}\text{Substitute}\ f(x+h)=\sqrt{x+h}\ \text{and}\ f(x)=\sqrt{x} \\ \text{into}\ {f}^{'}(x)=\underset{h\to 0}{\text{lim}}\frac{f(x+h)-f(x)}{h}.\end{array} \\ & =\underset{h\to 0}{\text{lim}}\frac{\sqrt{x+h}-\sqrt{x}}{h}\cdot \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} & & & \begin{array}{l}\text{Multiply numerator and denominator by} \\ \sqrt{x+h}+\sqrt{x}\ \text{without distributing in the} \\ \text{denominator.}\end{array} \\ & =\underset{h\to 0}{\text{lim}}\frac{h}{h(\sqrt{x+h}+\sqrt{x})} & & & \text{Multiply the numerators and simplify.} \\ & =\underset{h\to 0}{\text{lim}}\frac{1}{(\sqrt{x+h}+\sqrt{x})} & & & \text{Cancel the}\ h. \\ & =\frac{1}{2\sqrt{x}} & & & \text{Evaluate the limit.}\end{array}\]
Find the derivative of the function \(f(x)={x}^{2}-2x.\)

Giải đáp

Follow the same procedure here, but without having to multiply by the conjugate.
\[\begin{array}{lllll}{f}^{'}(x) & =\underset{h\to 0}{\text{lim}}\frac{({(x+h)}^{2}-2(x+h))-({x}^{2}-2x)}{h} & & & \begin{array}{l}\text{Substitute}\ f(x+h)={(x+h)}^{2}-2(x+h)\ \text{and} \\ f(x)={x}^{2}-2x\ \text{into} \\ {f}^{'}(x)=\underset{h\to 0}{\text{lim}}\frac{f(x+h)-f(x)}{h}.\end{array} \\ & =\underset{h\to 0}{\text{lim}}\frac{{x}^{2}+2xh+{h}^{2}-2x-2h-{x}^{2}+2x}{h} & & & \text{Expand}\ {(x+h)}^{2}-2(x+h). \\ & =\underset{h\to 0}{\text{lim}}\frac{2xh-2h+{h}^{2}}{h} & & & \text{Simplify.} \\ & =\underset{h\to 0}{\text{lim}}\frac{h(2x-2+h)}{h} & & & \text{Factor out}\ h\ \text{from the numerator.} \\ & =\underset{h\to 0}{\text{lim}}(2x-2+h) & & & \text{Cancel the common factor of}\ h. \\ & =2x-2 & & & \text{Evaluate the limit.}\end{array}\]
Find the derivative of \(f(x)={x}^{2}.\)

Giải đáp

\({f}^{'}(x)=2x\)
Use the following graph of \(f(x)\) to sketch a graph of \({f}^{'}(x).\)

Giải đáp

The solution is shown in the following graph. Observe that \(f(x)\) is increasing and \({f}^{'}(x)>0\) on \((-2,3).\) Also, \(f(x)\) is decreasing and \({f}^{'}(x)<0\) on \((\text{-}\infty ,-2)\) and on \((3,\text{+}\infty ).\) Also note that \(f(x)\) has horizontal tangents at \(-2\) and \(3,\) and \({f}^{'}(-2)=0\) and \({f}^{'}(3)=0.\)
Sketch the graph of \(f(x)={x}^{2}-4.\) On what interval is the graph of \({f}^{'}(x)\) above the \(x\)-axis?

Giải đáp

\((0,\text{+}\infty )\)
A toy company wants to design a track for a toy car that starts out along a parabolic curve and then converts to a straight line (). The function that describes the track is to have the form \(f(x)=\{\begin{array}{l}\frac{1}{10}{x}^{2}+bx+c\ \text{if}\ x<-10 \\ -\frac{1}{4}x+\frac{5}{2}\ \text{if}\ x\ge -10\end{array}\) where \(x\) and \(f(x)\) are in inches. For the car to move smoothly along the track, the function \(f(x)\) must be both continuous and differentiable at \(-10.\) Find values of \(b\) and \(c\) that make \(f(x)\) both continuous and differentiable.

Giải đáp

For the function to be continuous at \(x=-10,\underset{x\to {-10}^{-}}{\text{lim}}f(x)=f(-10).\) Thus, since
\[\underset{x\to \text{-}{10}^{-}}{\text{lim}}f(x)=\frac{1}{10}{(-10)}^{2}-10b+c=10-10b+c\]
and \(f(-10)=5,\) we must have \(10-10b+c=5.\) Equivalently, we have \(c=10b-5.\)
For the function to be differentiable at \(-10,\)
\[{f}^{'}(-10)=\underset{x\to \text{-}10}{\text{lim}}\frac{f(x)-f(-10)}{x+10}\]
must exist. Since \(f(x)\) is defined using different rules on the right and the left, we must evaluate this limit from the right and the left and then set them equal to each other:
\[\begin{array}{lllll}\underset{x\to \text{-}{10}^{-}}{\text{lim}}\frac{f(x)-f(-10)}{x+10} & =\underset{x\to \text{-}{10}^{-}}{\text{lim}}\frac{\frac{1}{10}{x}^{2}+bx+c-5}{x+10} & & & \\ & =\underset{x\to \text{-}{10}^{-}}{\text{lim}}\frac{\frac{1}{10}{x}^{2}+bx+(10b-5)-5}{x+10} & & & \text{Substitute}\ c=10b-5. \\ & =\underset{x\to \text{-}{10}^{-}}{\text{lim}}\frac{{x}^{2}-100+10bx+100b}{10(x+10)} & & & \text{Multiply numerator and denominator by 10.} \\ & =\underset{x\to \text{-}{10}^{-}}{\text{lim}}\frac{(x+10)(x-10+10b)}{10(x+10)} & & & \text{Factor by grouping.} \\ & =b-2. & & & \end{array}\]
We also have
\[\begin{array}{ll}\underset{x\to \text{-}{10}^{+}}{\text{lim}}\frac{f(x)-f(-10)}{x+10} & =\underset{x\to \text{-}{10}^{+}}{\text{lim}}\frac{-\frac{1}{4}x+\frac{5}{2}-5}{x+10} \\ & =\underset{x\to \text{-}{10}^{+}}{\text{lim}}\frac{\text{-}(x+10)}{4(x+10)} \\ & =-\frac{1}{4}.\end{array}\]
This gives us \(b-2=-\frac{1}{4}.\) Thus \(b=\frac{7}{4}\) and \(c=10(\frac{7}{4})-5=\frac{25}{2}.\)
Find values of \(a\) and \(b\) that make \(f(x)=\{\begin{array}{l}ax+b\ \text{if}\ x<3 \\ {x}^{2}\ \text{if}\ x\ge 3\end{array}\) both continuous and differentiable at \(3.\)

Giải đáp

\(a=6\) and \(b=-9\)
For \(f(x)=2{x}^{2}-3x+1,\) find \({f}^{″}(x).\)

Giải đáp

First find \({f}^{'}(x).\)
\[\begin{array}{lllll}{f}^{'}(x) & =\underset{h\to 0}{\text{lim}}\frac{(2{(x+h)}^{2}-3(x+h)+1)-(2{x}^{2}-3x+1)}{h} & & & \begin{array}{l}\text{Substitute}\ f(x)=2{x}^{2}-3x+1 \\ \text{and} \\ f(x+h)=2{(x+h)}^{2}-3(x+h)+1 \\ \text{into}\ {f}^{'}(x)=\underset{h\to 0}{\text{lim}}\frac{f(x+h)-f(x)}{h}.\end{array} \\ & =\underset{h\to 0}{\text{lim}}\frac{4xh+2{h}^{2}-3h}{h} & & & \text{Simplify the numerator.} \\ & =\underset{h\to 0}{\text{lim}}(4x+2h-3) & & & \begin{array}{l}\text{Factor out the}\ h\ \text{in the numerator} \\ \text{and cancel with the}\ h\ \text{in the} \\ \text{denominator.}\end{array} \\ & =4x-3 & & & \text{Take the limit.}\end{array}\]
Next, find \({f}^{″}(x)\) by taking the derivative of \({f}^{'}(x)=4x-3.\)
\[\begin{array}{lllll}{f}^{″}(x) & =\underset{h\to 0}{\text{lim}}\frac{{f}^{'}(x+h)-{f}^{'}(x)}{h} & & & \begin{array}{l}\text{Use}\ {f}^{'}(x)=\underset{h\to 0}{\text{lim}}\frac{f(x+h)-f(x)}{h}\ \text{with}\ {f}^{'}(x)\ \text{in} \\ \text{place of}\ f(x).\end{array} \\ & =\underset{h\to 0}{\text{lim}}\frac{(4(x+h)-3)-(4x-3)}{h} & & & \begin{array}{l}\text{Substitute}\ {f}^{'}(x+h)=4(x+h)-3\ \text{and} \\ {f}^{'}(x)=4x-3.\end{array} \\ & =\underset{h\to 0}{\text{lim}}4 & & & \text{Simplify.} \\ & =4 & & & \text{Take the limit.}\end{array}\]
Find \({f}^{″}(x)\) for \(f(x)={x}^{2}.\)

Giải đáp

\({f}^{″}(x)=2\)
The position of a particle along a coordinate axis at time \(t\) (in seconds) is given by \(s(t)=3{t}^{2}-4t+1\) (in meters). Find the function that describes its acceleration at time \(t.\)

Giải đáp

Since \(v(t)={s}^{'}(t)\) and \(a(t)={v}^{'}(t)={s}^{″}(t),\) we begin by finding the derivative of \(s(t):\)
\[\begin{array}{ll}{s}^{'}(t) & =\underset{h\to 0}{\text{lim}}\frac{s(t+h)-s(t)}{h} \\ & =\underset{h\to 0}{\text{lim}}\frac{3{(t+h)}^{2}-4(t+h)+1-(3{t}^{2}-4t+1)}{h} \\ & =6t-4.\end{array}\]
Next,
\[\begin{array}{ll}{s}^{″}(t) & =\underset{h\to 0}{\text{lim}}\frac{{s}^{'}(t+h)-{s}^{'}(t)}{h} \\ & =\underset{h\to 0}{\text{lim}}\frac{6(t+h)-4-(6t-4)}{h} \\ & =6.\end{array}\]
Thus, \(a=6{\ \text{m/s}}^{2}.\)
For \(s(t)={t}^{3},\) find \(a(t).\)

Giải đáp

\(a(t)=6t\)
\(f(x)=2-3x\)

Giải đáp

\(-3\)
\(f(x)=\frac{2x}{7}+1\)
\(f(x)=4{x}^{2}\)

Giải đáp

\(8x\)
\(f(x)=5x-{x}^{2}\)
\(f(x)=\sqrt{2x}\)

Giải đáp

\(\frac{1}{\sqrt{2x}}\)
\(f(x)=\sqrt{x-6}\)
\(f(x)=\frac{9}{x}\)

Giải đáp

\(\frac{-9}{{x}^{2}}\)
\(f(x)=x+\frac{1}{x}\)
\(f(x)=\frac{1}{\sqrt{x}}\)

Giải đáp

\(\frac{-1}{2{x}^{3\text{/}2}}\)
\(\underset{h\to 0}{\text{lim}}\frac{{(1+h)}^{2\text{/}3}-1}{h}\)
\(\underset{h\to 0}{\text{lim}}\frac{[3{(2+h)}^{2}+2]-14}{h}\)

Giải đáp

\(f(x)=3{x}^{2}+2,a=2\)
\(\underset{h\to 0}{\text{lim}}\frac{\text{cos}\ (\pi +h)+1}{h}\)
\(\underset{h\to 0}{\text{lim}}\frac{{(2+h)}^{4}-16}{h}\)

Giải đáp

\(f(x)={x}^{4},a=2\)
\(\underset{h\to 0}{\text{lim}}\frac{[2{(3+h)}^{2}-(3+h)]-15}{h}\)
\(\underset{h\to 0}{\text{lim}}\frac{{e}^{h}-1}{h}\)

Giải đáp

\(f(x)={e}^{x},a=0\)
\(f(x)=\{\begin{array}{l}2\sqrt{x},0\le x\le 1 \\ 3x-1,x>1\end{array}\)
\(f(x)=\{\begin{array}{l}3,x<1 \\ 3x,x\ge 1\end{array}\)

Giải đáp

a.

b. \(\underset{h\to {0}^{-}}{\text{lim}}\frac{f(1+h)-f(1)}{h}=\underset{h\to {0}^{-}}{\text{lim}}\frac{3-3}{h}=0\)
\(\underset{h\to {0}^{+}}{\text{lim}}\frac{f(1+h)-f(1)}{h}=\underset{h\to {0}^{+}}{\text{lim}}\frac{3(1+h)-3(1)}{h}=\underset{h\to {0}^{+}}{\text{lim}}\frac{3h}{h}=3\)
\(\underset{h\to {0}^{-}}{\text{lim}}\frac{f(1+h)-f(1)}{h}\ne \underset{h\to {0}^{+}}{\text{lim}}\frac{f(1+h)-f(1)}{h}\), so \(\underset{h\to 0}{\text{lim}}\frac{f(1+h)-f(1)}{h}\) does not exist.
\(f(x)=\{\begin{array}{l}-{x}^{2}+2,x\le 1 \\ x,x>1\end{array}\)
\(f(x)=\{\begin{array}{l}2x,x\le 1 \\ \frac{2}{x},x>1\end{array}\)

Giải đáp

a.

b. \(\underset{h\to {0}^{-}}{\text{lim}}\frac{f(1+h)-f(1)}{h}=\underset{h\to {0}^{-}}{\text{lim}}\frac{2(1+h)-2}{h}=\underset{h\to {0}^{-}}{\text{lim}}\frac{2h}{h}=\underset{h\to {0}^{-}}{\text{lim}}2=2\)
\(\begin{array}{ll}\underset{h\to {0}^{+}}{\text{lim}}\frac{f(1+h)-f(1)}{h} & =\underset{h\to {0}^{+}}{\text{lim}}\frac{\frac{2}{1+h}-\frac{2}{1}}{h} \\ & =\underset{h\to {0}^{+}}{\text{lim}}\frac{\frac{2-2(1+h)}{1+h}}{h} \\ & =\underset{h\to {0}^{+}}{\text{lim}}\frac{-2h}{h(1+h)} \\ & =\underset{h\to {0}^{+}}{\text{lim}}\frac{-2}{1+h} \\ & =-2\end{array}\)
\(\underset{h\to {0}^{-}}{\text{lim}}\frac{f(1+h)-f(1)}{h}\ne \underset{h\to {0}^{+}}{\text{lim}}\frac{f(1+h)-f(1)}{h}\), so \(\underset{h\to 0}{\text{lim}}\frac{f(1+h)-f(1)}{h}\) does not exist.
Use the graph of \(f(x)\) shown to evaluate a. \({f}^{'}(-0.5),\) b. \({f}^{'}(0),\) c. \({f}^{'}(1),\) d. \({f}^{'}(2),\) and e. \({f}^{'}(3),\) if it exists.
\(f(x)=2-3x\)

Giải đáp

\(0\)
\(f(x)=4{x}^{2}\)
\(f(x)=x+\frac{1}{x}\)

Giải đáp

\(\frac{2}{{x}^{3}}\)
[T] \(f(x)=-\frac{5}{x}\)
[T] \(f(x)=3{x}^{2}+2x+4.\)

Giải đáp

\({f}^{'}(x)=6x+2\)
[T] \(f(x)=\sqrt{x}+3x\)
[T] \(f(x)=\frac{1}{\sqrt{2x}}\)

Giải đáp

\({f}^{'}(x)=-\frac{1}{{(2x)}^{3\text{/}2}}\)
[T] \(f(x)=1+x+\frac{1}{x}\)
[T] \(f(x)={x}^{3}+1\)

Giải đáp

\({f}^{'}(x)=3{x}^{2}\)

Symbols used here

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

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

\infty

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

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

derivative
Instantaneous rate of change; slope of the graph.

\leq,\ \geq

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

\sup,\ \inf

supremum, infimum
Least upper bound, greatest lower bound.

\pi

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

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

\sin,\ \cos,\ \tan

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

\log_b x,\ \ln x

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

\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 Derivative as a Function

Define the derivative function of a given function.
Graph a derivative function from the graph of a given function.
State the connection between derivatives and continuity.
Describe three conditions for when a function does not have a derivative.
Explain the meaning of a higher-order derivative.
We observe that if a function is not continuous, it cannot be differentiable, since every differentiable function must be continuous. However, if a function is continuous, it may still fail to be differentiable.
We saw that
As we saw in the example of

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.

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

More in 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