3.2: The Derivative as a Function
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As we have seen, the derivative of a function at a given point gives us the rate of change or slope of the tangent line to the function at that point. If we differentiate a position function at a given time, we obtain the velocity at that time. It seems reasonable to conclude that knowing the derivative of the function at every point would produce valuable information about the behavior of the function. However, the process of finding the derivative at even a handful of values using the techniques of the preceding section would quickly become quite tedious. In this section we define the derivative function and learn a process for finding it.
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.
Definition: derivative function
Let \(f\) be a function. The derivative function, denoted by f′, is the function whose domain consists of those values of \(x\) such that the following limit exists:
\[f′(x)=\lim_{h→0}\frac{f(x+h)−f(x)}{h}.\]
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 Equation to find the derivative of a function.
Example \(\PageIndex{1}\): Finding the Derivative of a SquareRoot Function
Find the derivative of \(f(x)=\sqrt{x}\).
Solution
Start directly with the definition of the derivative function.
\(f′(x)=\displaystyle \lim_{h→0}\frac{\sqrt{x+h}−\sqrt{x}}{h}\)  Substitute\(f(x+h)=\sqrt{x+h}\) and \(f(x)=\sqrt{x}\) into \(f′(x)= \displaystyle \lim_{h→0}\frac{f(x+h)−f(x)}{h}\). 
\(=\displaystyle\lim_{h→0}\frac{\sqrt{x+h}−\sqrt{x}}{h}⋅\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\)  Multiply numerator and denominator by \(\sqrt{x+h}+\sqrt{x}\) without distributing in the denominator. 
\(=\displaystyle\lim_{h→0}\frac{h}{h(\sqrt{x+h}+\sqrt{x}})\)  Multiply the numerators and simplify. 
\(=\displaystyle\lim_{h→0}\frac{1}{(\sqrt{x+h}+\sqrt{x})}\)  Cancel the \(h\). 
\(=\frac{1}{2\sqrt{x}}\)  Evaluate the limit 
Example \(\PageIndex{2}\): Finding the Derivative of a Quadratic Function
Find the derivative of the function \(f(x)=x^2−2x\).
Solution
Follow the same procedure here, but without having to multiply by the conjugate.
\(f′(x)=\displaystyle\lim_{h→0}\frac{((x+h)^2−2(x+h))−(x^2−2x)}{h}\)  Substitute \(f(x+h)=(x+h)^2−2(x+h)\) and \(f(x)=x^2−2x\) into \(f′(x)= \displaystyle \lim_{h→0}\frac{f(x+h)−f(x)}{h}\) 
\(=\displaystyle\lim_{h→0}\frac{x^2+2xh+h^2−2x−2h−x^2+2x}{h}\)  Expand \((x+h)^2−2(x+h)\). 
\(=\displaystyle\lim_{h→0}\frac{2xh−2h+h^2}{h}\)  Simplify 
\(=\displaystyle\lim_{h→0}\frac{h(2x−2+h)}{h}\)  Factor out \(h\) from the numerator 
\(=\displaystyle\lim_{h→0}(2x−2+h)\)  Cancel the common factor of \(h\) 
\(=2x−2\)  Evaluate the limit 
\(\PageIndex{1}\)
Find the derivative of \(f(x)=x^2\).
 Hint

Use Equation and follow the example.
 Answer

\(f′(x)=2x\)
We use a variety of different notations to express the derivative of a function. In Example 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 \(f(x)\):
\(f′(x), \frac{dy}{dx},y′,\frac{d}{dx}(f(x))\).
In place of \(f′(a)\) we may also use \(\frac{dy}{dx}∣x=a\) Use of the \(\frac{dy}{dx}\) notation (called Leibniz notation) is quite common in engineering and physics. To understand this notation better, recall that the derivative of a function at a point is the limit of the slopes of secant lines as the secant lines approach the tangent line. The slopes of these secant lines are often expressed in the form \(\frac{Δy}{Δx}\) where \(Δy\) is the difference in the \(y\) values corresponding to the difference in the \(x\) values, which are expressed as \(Δx\) (Figure). Thus the derivative, which can be thought of as the instantaneous rate of change of y with respect to \(x\), is expressed as
\(\frac{dy}{dx}=\displaystyle \lim_{Δx→0}\frac{Δy}{Δx}\).
Figure \(\PageIndex{1}\): The derivative is expressed as \(\frac{dy}{dx}=\displaystyle\lim_{Δx→0}\frac{Δy}{Δx}\).
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 Example we found that for \(f(x)=\sqrt{x},f′(x)=1/2\sqrt{x}\). If we graph these functions on the same axes, as in Figure, 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 \(\displaystyle \lim_{x→0^+}f′(x)=+∞\), corresponding to a vertical tangent to \(f(x)\) at \(0\).
Figure \(\PageIndex{2}\): The derivative \(f′(x)\) is positive everywhere because the function \(f(x)\) is increasing.
In Example we found that for \(f(x)=x^2−2x,f′(x)=2x−2\). The graphs of these functions are shown in Figure. 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 \(f′(1)=0\).
Figure \(\PageIndex{3}\): The derivative \(f′(x)<0\) where the function \(f(x)\) is decreasing and \(f′(x)>0\) where \(f(x)\) is increasing. The derivative is zero where the function has a horizontal tangent
Example \(\PageIndex{3}\): Sketching a Derivative Using a Function
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 \((−∞,−2)\) and on \((3,+∞)\). Also note that \(f(x)\) has horizontal tangents at \(–2\) and \(3\), and \(f′(−2)=0\) and \(f′(3)=0\).
\(\PageIndex{2}\)
Sketch the graph of \(f(x)=x^2−4\). On what interval is the graph of \(f′(x)\) above the \(x\)axis?
 Hint

The graph of \(f′(x)\) is positive where \(f(x)\) is increasing.
 Answer

(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.
Differentiability Implies Continuity
Let \(f(x)\) be a function and \(a\) be in its domain. If \(f(x)\) is differentiable at \(a\), then \(f\) is continuous at \(a\).
Proof
If \(f(x)\) is differentiable at \(a\), then \(f′(a)\) exists and
\(f′(a)=\displaystyle \lim_{x→a}\frac{f(x)−f(a)}{x−a}\).
We want to show that \(f(x)\) is continuous at \(a\) by showing that \(\displaystyle \lim_{x→a}f(x)=f(a).\) Thus,
\(\displaystyle \lim_{x→a}f(x)=\lim_{x→a}(f(x)−f(a)+f(a))\)
\(=\displaystyle \lim_{x→a}(\frac{f(x)−f(a)}{x−a}⋅(x−a)+f(a))\) Multiply and divide \(f(x)−f(a)\) by \(x−a\).
\(=(\displaystyle \lim_{x→a}\frac{f(x)−f(a)}{x−a})⋅(\displaystyle \lim_{x→a}(x−a))+\displaystyle \lim_{x→a}f(a)\)
\(=f'(a)⋅0+f(a)\)
\(=f(a).\)
Therefore, since \(f(a)\) is defined and \(\displaystyle \lim_{x→a}f(x)=f(a)\), we conclude that \(f\) is continuous at \(a\).
□
We have just proven that differentiability implies continuity, but now we consider whether continuity implies differentiability. To determine an answer to this question, we examine the function \(f(x)=x\). This function is continuous everywhere; however, \(f′(0)\) is undefined. This observation leads us to believe that continuity does not imply differentiability. Let’s explore further. For \(f(x)=x\),
\(f′(0)=\displaystyle \lim_{x→0}\frac{f(x)−f(0)}{x−0}=\displaystyle \lim_{x→0}\frac{x−0}{x−0}=\displaystyle \lim_{x→0}\frac{x}{x}\).
This limit does not exist because
\(\displaystyle \lim_{x→0^−}\frac{x}{x}=−1\) and \(\displaystyle \lim_{x→0^+}\frac{x}{x}=1\).
See Figure \(\PageIndex{4}\).
Figure \(\PageIndex{4}\): The function \(f(x)=x\) is continuous at \(0\) but is not differentiable at \(0\).
Let’s consider some additional situations in which a continuous function fails to be differentiable. Consider the function \(f(x)=\sqrt[3]{x}\):
\(f′(0)=\displaystyle \lim_{x→0}\frac{\sqrt[3]{x}−0}{x−0}=\displaystyle \lim_{x→0}\frac{1}{\sqrt[3]{x^2}}=+∞\).
Thus \(f′(0)\) does not exist. A quick look at the graph of \(f(x)=\sqrt[3]{x}\) clarifies the situation. The function has a vertical tangent line at \(0\) (Figure).
Figure \(\PageIndex{5}\): The function \(f(x)=\sqrt[3]{x}\) has a vertical tangent at \(x=0\). It is continuous at \(0\) but is not differentiable at \(0\).
The function \(f(x)=\begin{cases} xsin(\frac{1}{x}) & \mbox{ if } x≠0\\0 \mbox{ if } x=0\end{cases}\) also has a derivative that exhibits interesting behavior at \(0\). We see that
\(f′(0)=\displaystyle \lim_{x→0}\frac{xsin(1/x)−0}{x−0}=\displaystyle \lim_{x→0}sin(\frac{1}{x})\).
This limit does not exist, essentially because the slopes of the secant lines continuously change direction as they approach zero (Figure).
Figure \(\PageIndex{6}\): The function \(f(x)=\begin{cases}xsin(\frac{1}{x})&ifx≠0\\0 & ifx=0\end{cases}\) is not differentiable at \(0\).
In summary:
 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 \(f(x)=x\) failed to be differentiable at \(0\) because the limit of the slopes of the tangent lines on the left and right were not the same. Visually, this resulted in a sharp corner on the graph of the function at \(0.\) From this we conclude that in order to be differentiable at a point, a function must be “smooth” at that point.
 As we saw in the example of \(f(x)=\sqrt[3]{x}\), a function fails to be differentiable at a point where there is a vertical tangent line.
 As we saw with \(f(x)=\begin{cases}xsin(\frac{1}{x})& \mbox{ if } x≠0\\0 & \mbox{ if } x=0\end{cases}\) a function may fail to be differentiable at a point in more complicated ways as well.
Example \(\PageIndex{4}\): A Piecewise Function that is Continuous and Differentiable
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 (Figure). The function that describes the track is to have the form \(f(x)=\begin{cases}\frac{1}{10}x^2+bx+c & \mbox{ if }x<−10\\−\frac{1}{4}x+\frac{5}{2} & \mbox{ if } x≥−10\end{cases}\) 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.
Figure \(\PageIndex{7}\): For the car to move smoothly along the track, the function must be both continuous and differentiable.
Solution
For the function to be continuous at \(x=−10\), \(\displaystyle \lim_{x→10^−}f(x)=f(−10)\). Thus, since
\(\displaystyle \lim_{x→−10^−}f(x)=\frac{1}{10}(−10)^2−10b+c=10−10b+c\)
and \(f(−10)=5\), we must have \(10−10b+c=5\). Equivalently, we have \(c=10b−5\).
For the function to be differentiable at \(−10\),
\(f′(10)=\displaystyle \lim_{x→−10}\frac{f(x)−f(−10)}{x+10}\)
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:
\(\displaystyle \lim_{x→−10^−}\frac{f(x)−f(−10)}{x+10}=\displaystyle \lim_{x→−10^−}\frac{\frac{1}{10}x^2+bx+c−5}{x+10}\)
\(=\displaystyle \lim_{x→−10^−}\frac{\frac{1}{10}x2+bx+(10b−5)−5}{x+10}\) Substitute \(c=10b−5\).
\(=\displaystyle \lim_{x→−10^−}\frac{x^2−100+10bx+100b}{10(x+10)}\)
\(=\displaystyle \lim_{x→−10^−}\frac{(x+10)(x−10+10b)}{10(x+10)}\) Factor by grouping
\(=b−2\).
We also have
\(\displaystyle \lim_{x→−10^+}\frac{f(x)−f(−10)}{x+10}\)
\(=\displaystyle \lim_{x→−10^+}\frac{−\frac{1}{4}x+\frac{5}{2}−5}{x+10}\)
\(=\displaystyle \lim_{x→−10^+}\frac{−(x+10)}{4(x+10)}\)
\(=−\frac{1}{4}\).
This gives us \(b−2=−\frac{1}{4}\). Thus \(b=\frac{7}{4}\) and \(c=10(\frac{7}{4})−5=\frac{25}{2}\).
\(\PageIndex{3}\)
Find values of a and b that make \(f(x)=\begin{cases}ax+b & \mbox{ if } x<3\\x^2 & \mbox{ if } x≥3\end{cases}\) both continuous and differentiable at \(3\).
 Hint

Use Example \(\PageIndex{4}\) as a guide.
 Answer

\(a=6\) and \(b=−9\)
HigherOrder 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 higherorder derivatives. The notation for the higherorder derivatives of \(y=f(x)\) can be expressed in any of the following forms:
\(f''(x),f'''(x),f^{(4)}(x),…,f^{(n)}(x)\)
\(y''(x),y'''(x),y^{(4)}(x),…,y^{(n)}(x)\)
\(\frac{d^2y}{dx^2},\frac{d^3y}{dy^3},\frac{d^4y}{dy^4},…,\frac{d^ny}{dy^n}.\)
It is interesting to note that the notation for \(\frac{d^2y}{dx^2}\) may be viewed as an attempt to express \(\frac{d}{dx}(\frac{dy}{dx})\) more compactly. Analogously, \(\frac{d}{dx}(\frac{d}{dx}(\frac{dy}{dx}))=\frac{d}{dx}(\frac{d^2y}{dx^2})=\frac{d^3y}{dx^3}\).
Example \(\PageIndex{5}\): Finding a Second Derivative
For \(f(x)=2x^2−3x+1\), find \(f''(x)\).
Solution
First find \(f′(x)\).
\(f′(x)=\displaystyle \lim_{h→0}\frac{(2(x+h)^2−3(x+h)+1)−(2x^2−3x+1)}{h}\)  Substitute \(f(x)=2x^2−3x+1\) and \(f(x+h)=2(x+h)^2−3(x+h)+1\) into \(f′(x)=\displaystyle \lim_{h→0}\frac{f(x+h)−f(x)}{h}.\) 
\(=\displaystyle \lim_{h→0}\frac{4xh+h^2−3h}{h}\)  Simplify the numerator. 
\(=\displaystyle \lim_{h→0}(4x+h−3)\)  Factor out the \(h\) in the numerator and cancel with the \(h\) in the denominator. 
\(=4x−3\)  Take the limit. 
Next, find \(f''(x)\) by taking the derivative of \(f′(x)=4x−3.\)
\(f''(x)=\displaystyle \lim_{h→0}\frac{f′(x+h)−f′(x)}{h}\)  Use \(f′(x)=\displaystyle \lim_{h→0}\frac{f(x+h)−f(x)}{h}\) with \(f ′(x)\) in place of \(f(x).\) 
\(=\displaystyle \lim_{h→0}\frac{(4(x+h)−3)−(4x−3)}{h}\)  Substitute \(f′(x+h)=4(x+h)−3\) and \(f′(x)=4x−3.\) 
\(=\displaystyle \lim_{h→0}4\)  Simplify. 
\(=4\)  Take the limit. 
\(\PageIndex{4}\)
Find \(f''(x)\) for \(f(x)=x^2\).
 Hint

We found \(f′(x)=2x\) in a previous checkpoint. Use Equation to find the derivative of \(f′(x)\)
 Answer

\(f''(x)=2\)
Example \(\PageIndex{6}\): Finding Acceleration
The position of a particle along a coordinate axis at time \(t\) (in seconds) is given by \(s(t)=3t^2−4t+1\) (in meters). Find the function that describes its acceleration at time \(t\).
Solution
Since \(v(t)=s′(t)\) and \(a(t)=v′(t)=s''(t)\), we begin by finding the derivative of \(s(t)\):
\(s′(t)=\displaystyle \lim_{h→0}\frac{s(t+h)−s(t)}{h}\)
\(=\displaystyle \lim_{h→0}\frac{3(t+h)^2−4(t+h)+1−(3t^2−4t+1)}{h}\)
\(=6t−4.\)
Next,
\(s''(t)=\displaystyle \lim_{h→0}\frac{s′(t+h)−s′(t)}{h}\)
\(=\displaystyle \lim_{h→0}\frac{6(t+h)−4−(6t−4)}{h}\)
\(=6.\)
Thus, \(a=6 m/s^2\).
\(\PageIndex{5}\)
For \(s(t)=t^3\), find \(a(t).\)
 Hint

Use Example \(\PageIndex{6}\) as a guide.
 Answer

\(a(t)=6t\)
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.
 Higherorder derivatives are derivatives of derivatives, from the second derivative to the \(nth\) derivative.
Key Equations
 The derivative function
\(f′(x)=\displaystyle \lim_{h→0}\frac{f(x+h)−f(x)}{h}\)
Glossary
 derivative function
 gives the derivative of a function at each point in the domain of the original function for which the derivative is defined
 differentiable at a
 a function for which \(f′(a)\) exists is differentiable at \(a\)
 differentiable on S
 a function for which \(f′(x)\) exists for each \(x\) in the open set \(S\) is differentiable on \(S\)
 differentiable function
 a function for which \(f′(x)\) exists is a differentiable function
 higherorder derivative
 a derivative of a derivative, from the second derivative to the nth derivative, is called a higherorder derivative
Contributors
Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CCBYSANC 4.0 license. Download for free at http://cnx.org.