OBJECTIVES

When you finish this section, you should be able to:

1. Define the derivative function (p. 154)
2. Graph the derivative function (p. 155)
3. Identify where a function has no derivative (p. 157)

Finding the Derivative of a Function at a Number \(c\)

Find the derivative of the function \(f( x) =x^{2}-5x\) at any real number \(c\) using form (2).

Solution Using form (2), we have \begin{eqnarray*} f^\prime ( c) &=&\lim\limits_{h\rightarrow 0}\dfrac{f(c+h) -f( c) }{h} = \lim\limits_{h\rightarrow 0}\dfrac{[ ( c+h) ^{2}-5( c+h) ] -( c^{2}-5c) }{h} \\[4pt] \notag &=&\lim\limits_{h\rightarrow 0}\dfrac{[ ( c^{2}+2ch+h^{2}) -5c-5h] -c^{2}+5c}{h}=\lim\limits_{h\rightarrow 0}\dfrac{2ch+h^{2}-5h}{ h} \\[4pt] \notag &=&\lim\limits_{h\rightarrow 0}\dfrac{h( 2c+h-5) }{h} =\lim\limits_{h\rightarrow 0}( 2c+h-5) =2c-5 \end{eqnarray*}

NOTE

Compare the solution and answer found in Example 1 to Example 4 on p. 149.

NOW WORK

Problem 9.

IN WORDS

In form (3) the derivative is the limit of a difference quotient.

spanDEFINITIONspan The Derivative Function \(f^\prime\)

The derivative function \(f^\prime\) of a function \(f\) is \[\bbox[5px, border:1px solid black, #F9F7ED]{ f^\prime ( x) =\lim\limits_{h\rightarrow 0}\dfrac{f( x+h) -f( x) }{h}}\tag{3} \]

provided the limit exists. If \(f\) has a derivative, then \(f\) is said to be differentiable.

Finding the Derivative Function

Differentiate \(f(x)=\sqrt{x}\) and determine the domain of \(f^\prime\).

Solution The domain of \(f\) is \(\{x|x ≥ 0\}\). To find the derivative of \(f\), we use form (3). Then \[ f^\prime ( x) =\lim\limits_{h\rightarrow 0}\dfrac{f( x+h) -f( x) }{h}=\lim\limits_{h\rightarrow 0}\dfrac{\sqrt{ x+h}-\sqrt{x}}{h} \]

We rationalize the numerator to find the limit. \begin{eqnarray*} f^\prime ( x) &=&\lim\limits_{h\rightarrow 0}\left[ \dfrac{\sqrt{ x+h}-\sqrt{x}}{h}\cdot \dfrac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x} } \right] =\lim\limits_{h\rightarrow 0}\dfrac{( x+h) -x}{ h( \sqrt{x+h}+\sqrt{x}) } \\[5pt] &=&\lim\limits_{h\rightarrow 0}\dfrac{h}{h( \sqrt{x+h}+\sqrt{x}) } =\lim\limits_{h\rightarrow 0}\dfrac{1}{\sqrt{x+h}+\sqrt{x}}=\dfrac{1}{2\sqrt{ x}} \end{eqnarray*}

Figure 8

The limit does not exist when \(x=0.\) But for all other \(x\) in the domain of \(f\), the limit does exist. So, the domain of the derivative function \(f^\prime (x) =\dfrac{1}{2\sqrt{x}}\) is \(\{ x|x>0\}\).

NOW WORK

Problem 15.

Interpreting the Derivative as a Rate of Change

Show that the rate of change of the area of a circle with respect to its radius is equal to its circumference.

Solution The area \(A\) = \(A(r)\) of a circle of radius \(r\) is \(A(r) =\pi r^{2}\). The derivative function gives the rate of change of the area with respect to the radius. \begin{eqnarray*} A' ( r) &=&\lim\limits_{h\rightarrow 0}\dfrac{A( r+h) -A( r) }{h}=\lim\limits_{h\rightarrow 0}\dfrac{\pi ( r+h) ^{2}-\pi r^{2}}{h}\\[5pt] &=&\lim\limits_{h\rightarrow 0}\dfrac{\pi (r^{2}+2rh+h^{2})-\pi r^{2}}{h}=\lim\limits_{h\rightarrow 0}\dfrac{\pi h( 2r+h) }{h}\\[5pt] &=&\lim\limits_{h\rightarrow 0}\pi ( 2r+h) =2\pi r \end{eqnarray*}

The rate of change of the area of a circle with respect to its radius is the circumference of the circle, \(2\pi r\).

NOW WORK

Problem 69.

Graphing a Function and Its Derivative

Find \(f^\prime\) if \(f( x) =x^{3}-1\). Then graph \(y=f( x)\) and \(y=f^\prime ( x)\) on the same set of coordinate axes.

Figure 9

Solution \(f( x) =x^{3}-1\) so \[ f( x+h) =( x+h) ^{3}-1=x^{3}+3hx^{2}+3h^{2}x+h^{3}-1 \] Using form (3), we find \begin{eqnarray*} f^\prime ( x) &=&\lim\limits_{h\rightarrow 0}\dfrac{f( x+h) -f( x) }{h}=\lim\limits_{h\rightarrow 0}\dfrac{( x^{3}+3hx^{2}+3h^{2}x+h^{3}-1) -( x^{3}-1) }{h}\\[7pt] &=&\lim\limits_{h\rightarrow 0}\dfrac{3hx^{2}+3h^{2}x+h^{3}}{h} =\lim\limits_{h\rightarrow 0}\dfrac{h( 3x^{2}+3hx+h^{2}) }{h}\\[7pt] &=&\lim\limits_{h\rightarrow 0}( 3x^{2}{+}3hx{+}h^{2}) =3x^{2} \end{eqnarray*}

The graphs of \(f\) and \(f^\prime\) are shown in Figure 9.

NOW WORK

Problem 19.

Graphing the Derivative Function

Figure 11

Use the graph of the function \(y=\) \(f(x)\), shown in Figure 11, to sketch the graph of the derivative function \(y=f^\prime (x)\).

Solution We begin by drawing tangent lines to the graph of \(f\) at the points shown in Figure 11. See the graph at the top of Figure 12. At the points \((-2, 3)\) and \(\left(\dfrac{3}{2},-2\right)\) the tangent lines are horizontal, so their slopes are 0. This means \(f^\prime ( -2) =0\) and \(f^\prime \left(\dfrac{3}{2}\right) =0,\) so the points \((-2, 0)\) and \(\left(\dfrac{3}{2},0\right)\) are on the graph of the derivative function. Now we estimate the slope of the tangent lines at the other selected points. For example, at the point \(( -4,-3) .\) the slope of the tangent line is positive and the line is rather steep. We estimate the slope to be close to \(6,\) and we plot the point \((-4,6)\) on the bottom graph of Figure 12. Continue the process and then connect the points with a smooth curve.

NOW WORK

Problem 29.

Identifying Where a Function Has No Derivative

Given the piecewise defined function \( f(x)=\left\{ \begin{array}{r@{\quad}rr} -2x^{2}+4 & \hbox{if} & x<1 \\ x^{2}+1 & \hbox{if} & x ≥ 1 \end{array} \right. ,\) determine if \(f^\prime\) (1) exists.

Solution We investigate the limit \[ \lim\limits_{x\rightarrow 1}\frac{f(x)-f(1)}{x-1}=\lim\limits_{x\rightarrow 1}\frac{f(x)-2}{x-1} \qquad {\color{#0066A7}{\hbox{\(f(1)=1^2+1=2\)}}} \]

If \(x < 1\), then \(f(x)=-2x^{2}+4\); if \(x ≥ 1\), then \(f(x)=x^{2}+1\). So, it is necessary to investigate the one-sided limits at 1. \begin{eqnarray*} \lim\limits_{x\rightarrow 1^{-}}\frac{f(x)-f(1)}{x-1}&=&\lim\limits_{x \rightarrow 1^{-}}\frac{(-2x^{2}+4)-2}{x-1} = \lim\limits_{x\rightarrow 1^{-}} \frac{-2(x^{2}-1)}{x-1}\\[5pt] &=&-2\lim\limits_{x\rightarrow 1^{-}}\frac{(x-1)(x+1)}{ x-1}=-2\lim\limits_{x\rightarrow 1^{-}}(x+1)=-4\\[5pt] \lim\limits_{x\rightarrow 1^{+}}\frac{f(x)-f(1)}{x-1}&=&\lim\limits_{x \rightarrow 1^{+}}\frac{(x^{2}+1)-2}{x-1} = \lim\limits_{x\rightarrow 1^{+}} \frac{(x-1)(x+1)}{x-1} = \lim\limits_{x\rightarrow 1^{+}}(x+1)=2 \end{eqnarray*} Since the one-sided limits are not equal, \(\lim\limits_{x\rightarrow 1} \dfrac{f( x) -f(1) }{x-1}\) does not exist, and so \( f^\prime (1)\) does not exist.

NOW WORK

Problem 39.

Showing That a Function Has No Derivative

Show that \(f( x) =\sqrt[5]{x+2}\) has no derivative at \(x=-2\).

Solution First we note that \(f(-2) =\sqrt[5]{-2+2}=0.\) Then we investigate the limit. \[ \lim\limits_{x\rightarrow -2}\dfrac{f( x) -f(-2) }{ x-(-2) }=\lim\limits_{x\rightarrow -2}\dfrac{\sqrt[5]{x+2}-0}{x+2 }=\lim\limits_{x\rightarrow -2}\dfrac{\sqrt[5]{x+2}}{x+2}=\lim\limits_{x \rightarrow -2}\dfrac{1}{\sqrt[5]{( x+2) ^{4}}}=\infty \]

Since the limit is infinite, the derivative of \(f\) does not exist at \(-2\).

NOW WORK

Problem 35.

THEOREM

If a function \(f\) has a derivative at a number \(c\), then \(f\) is continuous at \(c\).

NEED TO REVIEW?

Continuity is discussed in Section 1.3, pp. 93-99.

PROOF

To show that \(f\) is continuous at \(c\), we need to verify that \(\lim\limits_{x\rightarrow c}f(x)=f(c)\). We begin by observing that if \( x\neq c\), then \[ f(x)-f(c)=\left[ \frac{f(x)-f(c)}{x-c}\right] (x-c) \]

We take the limit of both sides as \(x\rightarrow c,\) and use the fact that the limit of a product equals the product of the limits, (we show later that each limit exists). \begin{eqnarray*} \lim\limits_{x\rightarrow c}[f(x)-f(c)]&=&\lim\limits_{x\rightarrow c}\left\{ \left[ \frac{f(x)-f(c)}{x-c}\right] (x-c)\right\}\\[6pt] &=&\left[ \lim\limits_{x \rightarrow c}\frac{f(x)-f(c)}{x-c}\right] \left[ \lim\limits_{x\rightarrow c}(x-c)\right] \end{eqnarray*}

Since \(f\) has a derivative at \(c\), we know that \[ \lim\limits_{x\rightarrow c}\frac{f(x)-f(c)}{x-c}=f^\prime (c) \] is a number. Also for any real number \(c\), \(\lim\limits_{x\rightarrow c}(x-c)=0\). So \[ \lim\limits_{x\rightarrow c}[f(x)-f(c)]=[f^\prime (c)]\left[ \lim\limits_{x\rightarrow c}(x-c)\right] =f^\prime (c)\cdot 0=0 \]

That is, \(\lim\limits_{x\rightarrow c}f(x)=f(c)\), so \(f\) is continuous at \(c\).

COROLLARY

If a function \(f\) is discontinuous at a number \(c\), then \(f\) has no derivative at \(c\).

IN WORDS

Differentiability implies continuity, but continuity does not imply differentiability.

Determining If a Heaviside Function Has a Derivative at 0

Determine if the Heaviside function \(u_{0}( t) =\left\{ \begin{array}{l@{\quad}ll} 0 & \hbox{if} & t < 0 \\ 1 & \hbox{if} & t ≥ 0 \end{array} \right.\) has a derivative at \(0.\)

Solution Since \(u_{0}( t)\) is discontinuous at 0, it has no derivative at 0.

NOW WORK

Problem 47.