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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 \(f(x)\text{:}\)

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

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 \(\underset{x\to {0}^{+}}{\text{lim}}{f}^{'}(x)=+\infty ,\) 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 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 \((3,+\infty ).\) 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,

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

For the following graphs,

  1. 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
  2. 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 \({f}^{'}(x).\)

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

이제 너 계산기는 이것을 해결하지 않지만, 그 조각은 계산 가능합니다. 아래의 하나를 시도하거나 자신의 것을 입력하십시오.

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먼저 종이에 각각을 시도해 보세요. 확인하기 위해 답을 드러내세요. 확인된 답은 각 단계의 솔버에서 열 수 있습니다.

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

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

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

  2. Find the derivative of the function \(f(x)={x}^{2}-2x.\)

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    Follow the same procedure here, but without having to multiply by the conjugate.

    \[\begin{array}{lllll}{f}^{'}(x) & =\underset{h\to 0}{\text{lim}}\frac{({(x+h)}^{2}-2(x+h))-({x}^{2}-2x)}{h} & & & \begin{array}{l}\text{Substitute}\ f(x+h)={(x+h)}^{2}-2(x+h)\ \text{and} \\ f(x)={x}^{2}-2x\ \text{into} \\ {f}^{'}(x)=\underset{h\to 0}{\text{lim}}\frac{f(x+h)-f(x)}{h}.\end{array} \\ & =\underset{h\to 0}{\text{lim}}\frac{{x}^{2}+2xh+{h}^{2}-2x-2h-{x}^{2}+2x}{h} & & & \text{Expand}\ {(x+h)}^{2}-2(x+h). \\ & =\underset{h\to 0}{\text{lim}}\frac{2xh-2h+{h}^{2}}{h} & & & \text{Simplify.} \\ & =\underset{h\to 0}{\text{lim}}\frac{h(2x-2+h)}{h} & & & \text{Factor out}\ h\ \text{from the numerator.} \\ & =\underset{h\to 0}{\text{lim}}(2x-2+h) & & & \text{Cancel the common factor of}\ h. \\ & =2x-2 & & & \text{Evaluate the limit.}\end{array}\]

  3. Find the derivative of \(f(x)={x}^{2}.\)

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

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

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    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 \((3,+\infty ).\) Also note that \(f(x)\) has horizontal tangents at \(-2\) and \(3,\) and \({f}^{'}(-2)=0\) and \({f}^{'}(3)=0.\)

  5. Sketch the graph of \(f(x)={x}^{2}-4.\) On what interval is the graph of \({f}^{'}(x)\) above the \(x\)-axis?

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    \((0,+\infty )\)

  6. 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.

    답을 드러내세요

    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 \(c=10(\frac{7}{4})-5=\frac{25}{2}.\)

  7. 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 \(3.\)

    답을 드러내세요

    \(a=6\) and \(b=-9\)

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

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    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 \({f}^{'}(x)=4x-3.\)

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

  9. Find \({f}^{″}(x)\) for \(f(x)={x}^{2}.\)

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    \({f}^{″}(x)=2\)

  10. 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 \(t.\)

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    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, \(a=6{\ \text{m/s}}^{2}.\)

  11. For \(s(t)={t}^{3},\) find \(a(t).\)

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    \(a(t)=6t\)

  12. \(f(x)=2-3x\)

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    \(-3\)

  13. \(f(x)=\frac{2x}{7}+1\)

  14. \(f(x)=4{x}^{2}\)

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    \(8x\)

  15. \(f(x)=5x-{x}^{2}\)

  16. \(f(x)=\sqrt{2x}\)

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    \(\frac{1}{\sqrt{2x}}\)

  17. \(f(x)=\sqrt{x-6}\)

  18. \(f(x)=\frac{9}{x}\)

    답을 드러내세요

    \(\frac{-9}{{x}^{2}}\)

  19. \(f(x)=x+\frac{1}{x}\)

  20. \(f(x)=\frac{1}{\sqrt{x}}\)

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    \(\frac{-1}{2{x}^{3\text{/}2}}\)

  21. \(\underset{h\to 0}{\text{lim}}\frac{{(1+h)}^{2\text{/}3}-1}{h}\)

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

    답을 드러내세요

    \(f(x)=3{x}^{2}+2,a=2\)

  23. \(\underset{h\to 0}{\text{lim}}\frac{\text{cos}\ (\pi +h)+1}{h}\)

  24. \(\underset{h\to 0}{\text{lim}}\frac{{(2+h)}^{4}-16}{h}\)

    답을 드러내세요

    \(f(x)={x}^{4},a=2\)

  25. \(\underset{h\to 0}{\text{lim}}\frac{[2{(3+h)}^{2}-(3+h)]-15}{h}\)

  26. \(\underset{h\to 0}{\text{lim}}\frac{{e}^{h}-1}{h}\)

    답을 드러내세요

    \(f(x)={e}^{x},a=0\)

  27. \(f(x)=\{\begin{array}{l}2\sqrt{x},0\le x\le 1 \\ 3x-1,x>1\end{array}\)

  28. \(f(x)=\{\begin{array}{l}3,x<1 \\ 3x,x\ge 1\end{array}\)

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    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.

  29. \(f(x)=\{\begin{array}{l}-{x}^{2}+2,x\le 1 \\ x,x>1\end{array}\)

  30. \(f(x)=\{\begin{array}{l}2x,x\le 1 \\ \frac{2}{x},x>1\end{array}\)

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    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.

  31. 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.

  32. \(f(x)=2-3x\)

    답을 드러내세요

    \(0\)

  33. \(f(x)=4{x}^{2}\)

  34. \(f(x)=x+\frac{1}{x}\)

    답을 드러내세요

    \(\frac{2}{{x}^{3}}\)

  35. [T] \(f(x)=-\frac{5}{x}\)

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

    답을 드러내세요

    \({f}^{'}(x)=6x+2\)

  37. [T] \(f(x)=\sqrt{x}+3x\)

  38. [T] \(f(x)=\frac{1}{\sqrt{2x}}\)

    답을 드러내세요

    \({f}^{'}(x)=-\frac{1}{{(2x)}^{3\text{/}2}}\)

  39. [T] \(f(x)=1+x+\frac{1}{x}\)

  40. [T] \(f(x)={x}^{3}+1\)

    답을 드러내세요

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

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어떻게: The Derivative as a Function

  1. Define the derivative function of a given function.
  2. Graph a derivative function from the graph of a given function.
  3. State the connection between derivatives and continuity.
  4. Describe three conditions for when a function does not have a derivative.
  5. Explain the meaning of a higher-order derivative.
  6. 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.
  7. We saw that
  8. As we saw in the example of

사람들이 묻는 질문

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.

이 페이지의 일부는 다음에서 변경되었습니다. OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0). 여기서 압축하고 다시 설명; 오류는 우리의. 소스가 비상업적 사용을 위해 라이선스되어 있기 때문에, 이 레슨은 무료이며, 아무것도 판매하지 않으며, 동일한 라이선스로 공유됩니다.

에 더 Calculus

LimitsDerivativesIntegralsDefinite integralsTaylor seriesSeries and sumsMaxima and minimaThe chain ruleImplicit differentiationRelated rates and optimisationIntegration techniques: substitution, parts, partial fractionsApplications of integration: area, volume, arc lengthInfinite series and convergence tests