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Study Guides > Calculus Volume 1

Derivatives of Exponential and Logarithmic Functions

Learning Objectives

  • Find the derivative of exponential functions.
  • Find the derivative of logarithmic functions.
  • Use logarithmic differentiation to determine the derivative of a function.

So far, we have learned how to differentiate a variety of functions, including trigonometric, inverse, and implicit functions. In this section, we explore derivatives of exponential and logarithmic functions. As we discussed in Introduction to Functions and Graphs, exponential functions play an important role in modeling population growth and the decay of radioactive materials. Logarithmic functions can help rescale large quantities and are particularly helpful for rewriting complicated expressions.

Derivative of the Exponential Function

Just as when we found the derivatives of other functions, we can find the derivatives of exponential and logarithmic functions using formulas. As we develop these formulas, we need to make certain basic assumptions. The proofs that these assumptions hold are beyond the scope of this course.

First of all, we begin with the assumption that the function B(x)=bx,b>0,B\left(x\right)={b}^{x},b>0, is defined for every real number and is continuous. In previous courses, the values of exponential functions for all rational numbers were defined—beginning with the definition of bn,{b}^{n}, where nn is a positive integer—as the product of bb multiplied by itself nn times. Later, we defined b0=1,b−n=1bn,{b}^{0}=1,{b}^{\text{−}n}=\frac{1}{{b}^{n}}, for a positive integer n,n, and {b}^{s\text{/}t}=\left(\sqrt[t]{b}{\right)}^{s} for positive integers ss and t.t. These definitions leave open the question of the value of br{b}^{r} where rr is an arbitrary real number. By assuming the continuity of B(x)=bx,b>0,B\left(x\right)={b}^{x},b>0, we may interpret br{b}^{r} as limx→rbx\underset{x\to r}{\text{lim}}{b}^{x} where the values of xx as we take the limit are rational. For example, we may view 4π{4}^{\pi } as the number satisfying

43<4π<44,43.1<4π<43.2,43.14<4π<43.15,43.141<4π<43.142,43.1415<4π<43.1416,….\begin{array}{l}{4}^{3}<{4}^{\pi }<{4}^{4},{4}^{3.1}<{4}^{\pi }<{4}^{3.2},{4}^{3.14}<{4}^{\pi }<{4}^{3.15},\\ {4}^{3.141}<{4}^{\pi }<{4}^{3.142},{4}^{3.1415}<{4}^{\pi }<{4}^{3.1416}\text{,}\text{…}.\end{array}

As we see in the following table, 4π≈77.88.{4}^{\pi }\approx 77.88.

Approximating a Value of 4π{4}^{\pi }
xx4x{4}^{x}xx4x{4}^{x}
43{4}^{3}6443.141593{4}^{3.141593}77.8802710486
43.1{4}^{3.1}73.516694719843.1416{4}^{3.1416}77.8810268071
43.14{4}^{3.14}77.708472601343.142{4}^{3.142}77.9242251944
43.141{4}^{3.141}77.816274123743.15{4}^{3.15}78.7932424541
43.1415{4}^{3.1415}77.870230952643.2{4}^{3.2}84.4485062895
43.14159{4}^{3.14159}77.879947154344{4}^{4}256

We also assume that for B(x)=bx,b>0,B\left(x\right)={b}^{x},b>0, the value B′(0){B}^{\prime }\left(0\right) of the derivative exists. In this section, we show that by making this one additional assumption, it is possible to prove that the function B(x)B\left(x\right) is differentiable everywhere.

We make one final assumption: that there is a unique value of b>0b>0 for which B′(0)=1.{B}^{\prime }\left(0\right)=1. We define ee to be this unique value, as we did in Introduction to Functions and Graphs. [link] provides graphs of the functions y=2x,y=3x,y=2.7x,y={2}^{x},y={3}^{x},y={2.7}^{x}, and y=2.8x.y={2.8}^{x}. A visual estimate of the slopes of the tangent lines to these functions at 0 provides evidence that the value of e lies somewhere between 2.7 and 2.8. The function E(x)=exE\left(x\right)={e}^{x} is called the natural exponential function. Its inverse, L(x)=logex=lnxL\left(x\right)={\text{log}}_{e}x=\text{ln}\phantom{\rule{0.1em}{0ex}}x is called the natural logarithmic function.

The graph of E(x)=exE\left(x\right)={e}^{x} is between y=2xy={2}^{x} and y=3x.y={3}^{x}.
The graphs of 3x, 2.8x, 2.7x, and 2x are shown. In quadrant I, their order from least to greatest is 2x, 2.7x, 2.8x, and 3x. In quadrant II, this order is reversed. All cross the y-axis at (0, 1).

For a better estimate of e,e, we may construct a table of estimates of B′(0){B}^{\prime }\left(0\right) for functions of the form B(x)=bx.B\left(x\right)={b}^{x}. Before doing this, recall that

B′(0)=limx→0bx−b0x−0=limx→0bx−1x≈bx−1x{B}^{\prime }\left(0\right)=\underset{x\to 0}{\text{lim}}\frac{{b}^{x}-{b}^{0}}{x-0}=\underset{x\to 0}{\text{lim}}\frac{{b}^{x}-1}{x}\approx \frac{{b}^{x}-1}{x}

for values of xx very close to zero. For our estimates, we choose x=0.00001x=0.00001 and x=−0.00001x=-0.00001 to obtain the estimate

b−0.00001−1−0.00001<B′(0)<b0.00001−10.00001.\frac{{b}^{-0.00001}-1}{-0.00001}<{B}^{\prime }\left(0\right)<\frac{{b}^{0.00001}-1}{0.00001}.

See the following table.

Estimating a Value of ee
bbb−0.00001−1−0.00001<B′(0)<b0.00001−10.00001\frac{{b}^{-0.00001}-1}{-0.00001}<{B}^{\prime }\left(0\right)<\frac{{b}^{0.00001}-1}{0.00001}bbb−0.00001−1−0.00001<B′(0)<b0.00001−10.00001\frac{{b}^{-0.00001}-1}{-0.00001}<{B}^{\prime }\left(0\right)<\frac{{b}^{0.00001}-1}{0.00001}
220.693145<B′(0)<0.693150.693145<{B}^{\prime }\left(0\right)<0.693152.71832.71831.000002<B′(0)<1.0000121.000002<{B}^{\prime }\left(0\right)<1.000012
2.72.70.993247<B′(0)<0.9932570.993247<{B}^{\prime }\left(0\right)<0.9932572.7192.7191.000259<B′(0)<1.0002691.000259<{B}^{\prime }\left(0\right)<1.000269
2.712.710.996944<B′(0)<0.9969540.996944<{B}^{\prime }\left(0\right)<0.9969542.722.721.000627<B′(0)<1.0006371.000627<{B}^{\prime }\left(0\right)<1.000637
2.7182.7180.999891<B′(0)<0.9999010.999891<{B}^{\prime }\left(0\right)<0.9999012.82.81.029614<B′(0)<1.0296251.029614<{B}^{\prime }\left(0\right)<1.029625
2.71822.71820.999965<B′(0)<0.9999750.999965<{B}^{\prime }\left(0\right)<0.999975331.098606<B′(0)<1.0986181.098606<{B}^{\prime }\left(0\right)<1.098618

The evidence from the table suggests that 2.7182<e<2.7183.2.7182<e<2.7183.

The graph of E(x)=exE\left(x\right)={e}^{x} together with the line y=x+1y=x+1 are shown in [link]. This line is tangent to the graph of E(x)=exE\left(x\right)={e}^{x} at x=0.x=0.

The tangent line to E(x)=exE\left(x\right)={e}^{x} at x=0x=0 has slope 1.
Graph of the function ex along with its tangent at (0, 1), x + 1.

Now that we have laid out our basic assumptions, we begin our investigation by exploring the derivative of B(x)=bx,b>0.B\left(x\right)={b}^{x},b>0. Recall that we have assumed that B′(0){B}^{\prime }\left(0\right) exists. By applying the limit definition to the derivative we conclude that

B′(0)=limh→0b0+h−b0h=limh→0bh−1h.{B}^{\prime }\left(0\right)=\underset{h\to 0}{\text{lim}}\frac{{b}^{0+h}-{b}^{0}}{h}=\underset{h\to 0}{\text{lim}}\frac{{b}^{h}-1}{h}.

Turning to B′(x),{B}^{\prime }\left(x\right), we obtain the following.

B′(x)=limh→0bx+h−bxhApply the limit definition of the derivative.=limh→0bxbh−bxhNote thatbx+h=bxbh.=limh→0bx(bh−1)hFactor outbx.=bxlimh→0bh−1hApply a property of limits.=bxB′(0)UseB′(0)=limh→0b0+h−b0h=limh→0bh−1h.\begin{array}{ccccc}\hfill {B}^{\prime }\left(x\right)& =\underset{h\to 0}{\text{lim}}\frac{{b}^{x+h}-{b}^{x}}{h}\hfill & & & \text{Apply the limit definition of the derivative.}\hfill \\ & =\underset{h\to 0}{\text{lim}}\frac{{b}^{x}{b}^{h}-{b}^{x}}{h}\hfill & & & \text{Note that}\phantom{\rule{0.2em}{0ex}}{b}^{x+h}={b}^{x}{b}^{h}.\hfill \\ & =\underset{h\to 0}{\text{lim}}\frac{{b}^{x}\left({b}^{h}-1\right)}{h}\hfill & & & \text{Factor out}\phantom{\rule{0.2em}{0ex}}{b}^{x}.\hfill \\ & ={b}^{x}\underset{h\to 0}{\text{lim}}\frac{{b}^{h}-1}{h}\hfill & & & \text{Apply a property of limits.}\hfill \\ & ={b}^{x}{B}^{\prime }\left(0\right)\hfill & & & \text{Use}\phantom{\rule{0.2em}{0ex}}{B}^{\prime }\left(0\right)=\underset{h\to 0}{\text{lim}}\frac{{b}^{0+h}-{b}^{0}}{h}=\underset{h\to 0}{\text{lim}}\frac{{b}^{h}-1}{h}.\hfill \end{array}

We see that on the basis of the assumption that B(x)=bxB\left(x\right)={b}^{x} is differentiable at 0,B(x)0,B\left(x\right) is not only differentiable everywhere, but its derivative is

B′(x)=bxB′(0).{B}^{\prime }\left(x\right)={b}^{x}{B}^{\prime }\left(0\right).

For E(x)=ex,E′(0)=1.E\left(x\right)={e}^{x},{E}^{\prime }\left(0\right)=1. Thus, we have E′(x)=ex.{E}^{\prime }\left(x\right)={e}^{x}. (The value of B′(0){B}^{\prime }\left(0\right) for an arbitrary function of the form B(x)=bx,b>0,B\left(x\right)={b}^{x},b>0, will be derived later.)

Derivative of the Natural Exponential Function

Let E(x)=exE\left(x\right)={e}^{x} be the natural exponential function. Then

E′(x)=ex.{E}^{\prime }\left(x\right)={e}^{x}.

In general,

ddx(eg(x))=eg(x)g′(x).\frac{d}{dx}\left({e}^{g\left(x\right)}\right)={e}^{g\left(x\right)}{g}^{\prime }\left(x\right).
Derivative of an Exponential Function

Find the derivative of f(x)=etan(2x).f\left(x\right)={e}^{\text{tan}\left(2x\right)}.

Using the derivative formula and the chain rule,

f′(x)=etan(2x)ddx(tan(2x))=etan(2x)sec2(2x)⋅2.\begin{array}{cc}\hfill {f}^{\prime }\left(x\right)& ={e}^{\text{tan}\phantom{\rule{0.1em}{0ex}}\left(2x\right)}\frac{d}{dx}\left(\text{tan}\phantom{\rule{0.1em}{0ex}}\left(2x\right)\right)\hfill \\ & ={e}^{\text{tan}\left(2x\right)}{\text{sec}}^{2}\left(2x\right)·2.\hfill \end{array}
Combining Differentiation Rules

Find the derivative of y=ex2x.y=\frac{{e}^{{x}^{2}}}{x}.

Use the derivative of the natural exponential function, the quotient rule, and the chain rule.

y′=(ex2⋅2)x⋅x−1⋅ex2x2Apply the quotient rule.=ex2(2x2−1)x2Simplify.\begin{array}{ccccc}\hfill {y}^{\prime }& =\frac{\left({e}^{{x}^{2}}·2\right)x·x-1·{e}^{{x}^{2}}}{{x}^{2}}\hfill & & & \text{Apply the quotient rule.}\hfill \\ & =\frac{{e}^{{x}^{2}}\left(2{x}^{2}-1\right)}{{x}^{2}}\hfill & & & \text{Simplify.}\hfill \end{array}

Find the derivative of h(x)=xe2x.h\left(x\right)=x{e}^{2x}.

h′(x)=e2x+2xe2x{h}^{\prime }\left(x\right)={e}^{2x}+2x{e}^{2x}

Hint

Don’t forget to use the product rule.

Applying the Natural Exponential Function

A colony of mosquitoes has an initial population of 1000. After tt days, the population is given by A(t)=1000e0.3t.A\left(t\right)=1000{e}^{0.3t}. Show that the ratio of the rate of change of the population, A′(t),{A}^{\prime }\left(t\right), to the population, A(t)A\left(t\right) is constant.

First find A′(t).{A}^{\prime }\left(t\right). By using the chain rule, we have A′(t)=300e0.3t.{A}^{\prime }\left(t\right)=300{e}^{0.3t}. Thus, the ratio of the rate of change of the population to the population is given by

A′(t)=300e0.3t1000e0.3t=0.3.{A}^{\prime }\left(t\right)=\frac{300{e}^{0.3t}}{1000{e}^{0.3t}}=0.3.

The ratio of the rate of change of the population to the population is the constant 0.3.

If A(t)=1000e0.3tA\left(t\right)=1000{e}^{0.3t} describes the mosquito population after tt days, as in the preceding example, what is the rate of change of A(t)A\left(t\right) after 4 days?

996

Hint

Find A′(4).{A}^{\prime }\left(4\right).

Derivative of the Logarithmic Function

Now that we have the derivative of the natural exponential function, we can use implicit differentiation to find the derivative of its inverse, the natural logarithmic function.

The Derivative of the Natural Logarithmic Function

If x>0x>0 and y=lnx,y=\text{ln}\phantom{\rule{0.1em}{0ex}}x, then

dydx=1x.\frac{dy}{dx}=\frac{1}{x}.

More generally, let g(x)g\left(x\right) be a differentiable function. For all values of xx for which g′(x)>0,{g}^{\prime }\left(x\right)>0, the derivative of h(x)=ln(g(x))h\left(x\right)=\text{ln}\phantom{\rule{0.1em}{0ex}}\left(g\left(x\right)\right) is given by

h′(x)=1g(x)g′(x).{h}^{\prime }\left(x\right)=\frac{1}{g\left(x\right)}{g}^{\prime }\left(x\right).

Proof

If x>0x>0 and y=lnx,y=\text{ln}\phantom{\rule{0.1em}{0ex}}x, then ey=x.{e}^{y}=x. Differentiating both sides of this equation results in the equation

eydydx=1.{e}^{y}\frac{dy}{dx}=1.

Solving for dydx\frac{dy}{dx} yields

dydx=1ey.\frac{dy}{dx}=\frac{1}{{e}^{y}}.

Finally, we substitute x=eyx={e}^{y} to obtain

dydx=1x.\frac{dy}{dx}=\frac{1}{x}.

We may also derive this result by applying the inverse function theorem, as follows. Since y=g(x)=lnxy=g\left(x\right)=\text{ln}\phantom{\rule{0.1em}{0ex}}x is the inverse of f(x)=ex,f\left(x\right)={e}^{x}, by applying the inverse function theorem we have

dydx=1f′(g(x))=1elnx=1x.\frac{dy}{dx}=\frac{1}{{f}^{\prime }\left(g\left(x\right)\right)}=\frac{1}{{e}^{\text{ln}\phantom{\rule{0.1em}{0ex}}x}}=\frac{1}{x}.

Using this result and applying the chain rule to h(x)=ln(g(x))h\left(x\right)=\text{ln}\phantom{\rule{0.1em}{0ex}}\left(g\left(x\right)\right) yields

h′(x)=1g(x)g′(x).{h}^{\prime }\left(x\right)=\frac{1}{g\left(x\right)}{g}^{\prime }\left(x\right).

□

The graph of y=lnxy=\text{ln}\phantom{\rule{0.1em}{0ex}}x and its derivative dydx=1x\frac{dy}{dx}=\frac{1}{x} are shown in [link].

The functiony=lnx\text{The function}\phantom{\rule{0.2em}{0ex}}y=\text{ln}\phantom{\rule{0.1em}{0ex}}x is increasing on (0,+∞).\left(0,\text{+}\infty \right). Its derivative y′=1xy\prime =\frac{1}{x} is greater than zero on (0,+∞).\left(0,\text{+}\infty \right).
Graph of the function ln x along with its derivative 1/x. The function ln x is increasing on (0, + ∞). Its derivative is decreasing but greater than 0 on (0, + ∞).
Taking a Derivative of a Natural Logarithm

Find the derivative of f(x)=ln(x3+3x−4).f\left(x\right)=\text{ln}\phantom{\rule{0.1em}{0ex}}\left({x}^{3}+3x-4\right).

Use [link] directly.

f′(x)=1x3+3x−4⋅(3x2+3)Useg(x)=x3+3x−4inh′(x)=1g(x)g′(x).=3x2+3x3+3x−4Rewrite.\begin{array}{ccccc}\hfill {f}^{\prime }\left(x\right)& =\frac{1}{{x}^{3}+3x-4}·\left(3{x}^{2}+3\right)\hfill & & & \text{Use}\phantom{\rule{0.2em}{0ex}}g\left(x\right)={x}^{3}+3x-4\phantom{\rule{0.2em}{0ex}}\text{in}\phantom{\rule{0.2em}{0ex}}{h}^{\prime }\left(x\right)=\frac{1}{g\left(x\right)}{g}^{\prime }\left(x\right).\hfill \\ & =\frac{3{x}^{2}+3}{{x}^{3}+3x-4}\hfill & & & \text{Rewrite.}\hfill \end{array}
Using Properties of Logarithms in a Derivative

Find the derivative of f(x)=ln(x2sinx2x+1).f\left(x\right)=\text{ln}\phantom{\rule{0.1em}{0ex}}\left(\frac{{x}^{2}\text{sin}\phantom{\rule{0.1em}{0ex}}x}{2x+1}\right).

At first glance, taking this derivative appears rather complicated. However, by using the properties of logarithms prior to finding the derivative, we can make the problem much simpler.

f(x)=ln(x2sinx2x+1)=2lnx+ln(sinx)−ln(2x+1)Apply properties of logarithms.f′(x)=2x+cotx−22x+1Apply sum rule andh′(x)=1g(x)g′(x).\begin{array}{cccccc}\hfill f\left(x\right)& =\hfill & \text{ln}\phantom{\rule{0.1em}{0ex}}\left(\frac{{x}^{2}\text{sin}\phantom{\rule{0.1em}{0ex}}x}{2x+1}\right)=2\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}x+\text{ln}\phantom{\rule{0.1em}{0ex}}\left(\text{sin}\phantom{\rule{0.1em}{0ex}}x\right)-\text{ln}\phantom{\rule{0.1em}{0ex}}\left(2x+1\right)\hfill & & & \text{Apply properties of logarithms.}\hfill \\ \hfill {f}^{\prime }\left(x\right)& =\hfill & \frac{2}{x}+\text{cot}\phantom{\rule{0.1em}{0ex}}x-\frac{2}{2x+1}\hfill & & & \text{Apply sum rule and}\phantom{\rule{0.2em}{0ex}}{h}^{\prime }\left(x\right)=\frac{1}{g\left(x\right)}{g}^{\prime }\left(x\right).\hfill \end{array}

Differentiate: f(x)=ln(3x+2)5.f\left(x\right)=\text{ln}{\left(3x+2\right)}^{5}.

f′(x)=153x+2{f}^{\prime }\left(x\right)=\frac{15}{3x+2}

Hint

Use a property of logarithms to simplify before taking the derivative.

Now that we can differentiate the natural logarithmic function, we can use this result to find the derivatives of y=logbxy=lo{g}_{b}x and y=bxy={b}^{x} for b>0,b≠1.b>0,b\ne 1.

Derivatives of General Exponential and Logarithmic Functions

Let b>0,b≠1,b>0,b\ne 1, and let g(x)g\left(x\right) be a differentiable function.

  1. If, y=logbx,y={\text{log}}_{b}x, then

    dydx=1xlnb.\frac{dy}{dx}=\frac{1}{x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}b}.

    More generally, if h(x)=logb(g(x)),h\left(x\right)={\text{log}}_{b}\left(g\left(x\right)\right), then for all values of x for which g(x)>0,g\left(x\right)>0,

    h′(x)=g′(x)g(x)lnb.{h}^{\prime }\left(x\right)=\frac{{g}^{\prime }\left(x\right)}{g\left(x\right)\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}b}.
  2. If y=bx,y={b}^{x}, then

    dydx=bxlnb.\frac{dy}{dx}={b}^{x}\text{ln}\phantom{\rule{0.1em}{0ex}}b.

    More generally, if h(x)=bg(x),h\left(x\right)={b}^{g\left(x\right)}, then

    h′(x)=bg(x)g″(x)lnb.{h}^{\prime }\left(x\right)={b}^{g\left(x\right)}g\text{″}\left(x\right)\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}b.

Proof

If y=logbx,y={\text{log}}_{b}x, then by=x.{b}^{y}=x. It follows that ln(by)=lnx.\text{ln}\phantom{\rule{0.1em}{0ex}}\left({b}^{y}\right)=\text{ln}\phantom{\rule{0.2em}{0ex}}x. Thus ylnb=lnx.y\phantom{\rule{0.2em}{0ex}}\text{ln}\phantom{\rule{0.2em}{0ex}}b=\text{ln}\phantom{\rule{0.2em}{0ex}}x. Solving for y,y, we have y=lnxlnb.y=\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}x}{\text{ln}\phantom{\rule{0.1em}{0ex}}b}. Differentiating and keeping in mind that lnb\text{ln}\phantom{\rule{0.1em}{0ex}}b is a constant, we see that

dydx=1xlnb.\frac{dy}{dx}=\frac{1}{x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}b}.

The derivative in [link] now follows from the chain rule.

If y=bx,y={b}^{x}, then lny=xlnb.\text{ln}\phantom{\rule{0.2em}{0ex}}y=x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}b. Using implicit differentiation, again keeping in mind that lnb\text{ln}\phantom{\rule{0.1em}{0ex}}b is constant, it follows that 1ydydx=lnb.\frac{1}{y}\phantom{\rule{0.2em}{0ex}}\frac{dy}{dx}=\text{ln}\phantom{\rule{0.1em}{0ex}}b. Solving for dydx\frac{dy}{dx} and substituting y=bx,y={b}^{x}, we see that

dydx=ylnb=bxlnb.\frac{dy}{dx}=y\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}b={b}^{x}\text{ln}\phantom{\rule{0.1em}{0ex}}b.

The more general derivative ([link]) follows from the chain rule.

□

Applying Derivative Formulas

Find the derivative of h(x)=3x3x+2.h\left(x\right)=\frac{{3}^{x}}{{3}^{x}+2}.

Use the quotient rule and [link].

h′(x)=3xln3(3x+2)−3xln3(3x)(3x+2)2Apply the quotient rule.=2⋅3xln3(3x+2)2Simplify.\begin{array}{ccccc}\hfill {h}^{\prime }\left(x\right)& =\frac{{3}^{x}\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}3\left({3}^{x}+2\right)-{3}^{x}\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}3\left({3}^{x}\right)}{{\left({3}^{x}+2\right)}^{2}}\hfill & & & \text{Apply the quotient rule.}\hfill \\ & =\frac{2·{3}^{x}\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}3}{{\left({3}^{x}+2\right)}^{2}}\hfill & & & \text{Simplify.}\hfill \end{array}
Finding the Slope of a Tangent Line

Find the slope of the line tangent to the graph of y=log2(3x+1)y={\text{log}}_{2}\left(3x+1\right) at x=1.x=1.

To find the slope, we must evaluate dydx\frac{dy}{dx} at x=1.x=1. Using [link], we see that

dydx=3ln2(3x+1).\frac{dy}{dx}=\frac{3}{\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}2\left(3x+1\right)}.

By evaluating the derivative at x=1,x=1, we see that the tangent line has slope

dydx∣x=1=34ln2=3ln16.\frac{dy}{dx}|\begin{array}{l}\\ {}_{x=1}\end{array}=\frac{3}{4\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}2}=\frac{3}{\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}16}.

Find the slope for the line tangent to y=3xy={3}^{x} at x=2.x=2.

9ln(3)9\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}\left(3\right)

Hint

Evaluate the derivative at x=2.x=2.

Logarithmic Differentiation

At this point, we can take derivatives of functions of the form y=(g(x))ny={\left(g\left(x\right)\right)}^{n} for certain values of n,n, as well as functions of the form y=bg(x),y={b}^{g\left(x\right)}, where b>0b>0 and b≠1.b\ne 1. Unfortunately, we still do not know the derivatives of functions such as y=xxy={x}^{x} or y=xπ.y={x}^{\pi }. These functions require a technique called logarithmic differentiation, which allows us to differentiate any function of the form h(x)=g(x)f(x).h\left(x\right)=g{\left(x\right)}^{f\left(x\right)}. It can also be used to convert a very complex differentiation problem into a simpler one, such as finding the derivative of y=x2x+1exsin3x.y=\frac{x\sqrt{2x+1}}{{e}^{x}{\text{sin}}^{3}x}. We outline this technique in the following problem-solving strategy.

Problem-Solving Strategy: Using Logarithmic Differentiation
  1. To differentiate y=h(x)y=h\left(x\right) using logarithmic differentiation, take the natural logarithm of both sides of the equation to obtain lny=ln(h(x)).\text{ln}\phantom{\rule{0.2em}{0ex}}y=\text{ln}\phantom{\rule{0.1em}{0ex}}\left(h\left(x\right)\right).
  2. Use properties of logarithms to expand ln(h(x))\text{ln}\phantom{\rule{0.1em}{0ex}}\left(h\left(x\right)\right) as much as possible.
  3. Differentiate both sides of the equation. On the left we will have 1ydydx.\frac{1}{y}\phantom{\rule{0.2em}{0ex}}\frac{dy}{dx}.
  4. Multiply both sides of the equation by yy to solve for dydx.\frac{dy}{dx}.
  5. Replace yy by h(x).h\left(x\right).
Using Logarithmic Differentiation

Find the derivative of y=(2x4+1)tanx.y={\left(2{x}^{4}+1\right)}^{\text{tan}\phantom{\rule{0.1em}{0ex}}x}.

Use logarithmic differentiation to find this derivative.

lny=ln(2x4+1)tanxStep 1. Take the natural logarithm of both sides.lny=tanxln(2x4+1)Step 2. Expand using properties of logarithms.1ydydx=sec2xln(2x4+1)+8x32x4+1⋅tanxStep 3. Differentiate both sides. Use theproduct rule on the right.dydx=y⋅(sec2xln(2x4+1)+8x32x4+1⋅tanx)Step 4. Multiply byyon both sides.dydx=(2x4+1)tanx(sec2xln(2x4+1)+8x32x4+1⋅tanx)Step 5. Substitutey=(2x4+1)tanx.\begin{array}{cccccc}\hfill \text{ln}\phantom{\rule{0.1em}{0ex}}y& =\hfill & \text{ln}{\left(2{x}^{4}+1\right)}^{\text{tan}\phantom{\rule{0.1em}{0ex}}x}\hfill & & & \text{Step 1. Take the natural logarithm of both sides.}\hfill \\ \hfill \text{ln}\phantom{\rule{0.1em}{0ex}}y& =\hfill & \text{tan}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}\left(2{x}^{4}+1\right)\hfill & & & \text{Step 2. Expand using properties of logarithms.}\hfill \\ \hfill \frac{1}{y}\phantom{\rule{0.2em}{0ex}}\frac{dy}{dx}& =\hfill & {\text{sec}}^{2}x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}\left(2{x}^{4}+1\right)+\frac{8{x}^{3}}{2{x}^{4}+1}·\text{tan}\phantom{\rule{0.1em}{0ex}}x\hfill & & & \begin{array}{c}\text{Step 3. Differentiate both sides. Use the}\hfill \\ \text{product rule on the right.}\hfill \end{array}\hfill \\ \hfill \frac{dy}{dx}& =\hfill & y·\left({\text{sec}}^{2}x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}\left(2{x}^{4}+1\right)+\frac{8{x}^{3}}{2{x}^{4}+1}·\text{tan}\phantom{\rule{0.1em}{0ex}}x\right)\hfill & & & \text{Step 4. Multiply by}\phantom{\rule{0.2em}{0ex}}y\phantom{\rule{0.2em}{0ex}}\text{on both sides.}\hfill \\ \hfill \frac{dy}{dx}& =\hfill & {\left(2{x}^{4}+1\right)}^{\text{tan}\phantom{\rule{0.1em}{0ex}}x}\left({\text{sec}}^{2}x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}\left(2{x}^{4}+1\right)+\frac{8{x}^{3}}{2{x}^{4}+1}·\text{tan}\phantom{\rule{0.1em}{0ex}}x\right)\hfill & & & \text{Step 5. Substitute}\phantom{\rule{0.2em}{0ex}}y={\left(2{x}^{4}+1\right)}^{\text{tan}\phantom{\rule{0.1em}{0ex}}x}.\hfill \end{array}
Using Logarithmic Differentiation

Find the derivative of y=x2x+1exsin3x.y=\frac{x\sqrt{2x+1}}{{e}^{x}{\text{sin}}^{3}x}.

This problem really makes use of the properties of logarithms and the differentiation rules given in this chapter.

lny=lnx2x+1exsin3xStep 1. Take the natural logarithm of both sides.lny=lnx+12ln(2x+1)−xlne−3lnsinxStep 2. Expand using properties of logarithms.1ydydx=1x+12x+1−1−3cosxsinxStep 3. Differentiate both sides.dydx=y(1x+12x+1−1−3cotx)Step 4. Multiply byyon both sides.dydx=x2x+1exsin3x(1x+12x+1−1−3cotx)Step 5. Substitutey=x2x+1exsin3x.\begin{array}{cccccc}\hfill \text{ln}\phantom{\rule{0.1em}{0ex}}y& =\hfill & \text{ln}\phantom{\rule{0.2em}{0ex}}\frac{x\sqrt{2x+1}}{{e}^{x}{\text{sin}}^{3}x}\hfill & & & \text{Step 1. Take the natural logarithm of both sides.}\hfill \\ \hfill \text{ln}\phantom{\rule{0.1em}{0ex}}y& =\hfill & \text{ln}\phantom{\rule{0.1em}{0ex}}x+\frac{1}{2}\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}\left(2x+1\right)-x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}e-3\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}x\hfill & & & \text{Step 2. Expand using properties of logarithms.}\hfill \\ \hfill \frac{1}{y}\phantom{\rule{0.2em}{0ex}}\frac{dy}{dx}& =\hfill & \frac{1}{x}+\frac{1}{2x+1}-1-3\frac{\text{cos}\phantom{\rule{0.1em}{0ex}}x}{\text{sin}\phantom{\rule{0.1em}{0ex}}x}\hfill & & & \text{Step 3. Differentiate both sides.}\hfill \\ \hfill \frac{dy}{dx}& =\hfill & y\left(\frac{1}{x}+\frac{1}{2x+1}-1-3\phantom{\rule{0.1em}{0ex}}\text{cot}\phantom{\rule{0.1em}{0ex}}x\right)\hfill & & & \text{Step 4. Multiply by}\phantom{\rule{0.2em}{0ex}}y\phantom{\rule{0.2em}{0ex}}\text{on both sides.}\hfill \\ \hfill \frac{dy}{dx}& =\hfill & \frac{x\sqrt{2x+1}}{{e}^{x}{\text{sin}}^{3}x}\left(\frac{1}{x}+\frac{1}{2x+1}-1-3\phantom{\rule{0.1em}{0ex}}\text{cot}\phantom{\rule{0.1em}{0ex}}x\right)\hfill & & & \text{Step 5. Substitute}\phantom{\rule{0.2em}{0ex}}y=\frac{x\sqrt{2x+1}}{{e}^{x}{\text{sin}}^{3}x}.\hfill \end{array}
Extending the Power Rule

Find the derivative of y=xry={x}^{r} where rr is an arbitrary real number.

The process is the same as in [link], though with fewer complications.

lny=lnxrStep 1. Take the natural logarithm of both sides.lny=rlnxStep 2. Expand using properties of logarithms.1ydydx=r1xStep 3. Differentiate both sides.dydx=yrxStep 4. Multiply byyon both sides.dydx=xrrxStep 5. Substitutey=xr.dydx=rxr−1Simplify.\begin{array}{cccccc}\hfill \text{ln}\phantom{\rule{0.1em}{0ex}}y& =\hfill & \text{ln}{x}^{r}\hfill & & & \text{Step 1. Take the natural logarithm of both sides.}\hfill \\ \hfill \text{ln}\phantom{\rule{0.1em}{0ex}}y& =\hfill & r\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}x\hfill & & & \text{Step 2. Expand using properties of logarithms.}\hfill \\ \hfill \frac{1}{y}\phantom{\rule{0.2em}{0ex}}\frac{dy}{dx}& =\hfill & r\frac{1}{x}\hfill & & & \text{Step 3. Differentiate both sides.}\hfill \\ \hfill \frac{dy}{dx}& =\hfill & y\frac{r}{x}\hfill & & & \text{Step 4. Multiply by}\phantom{\rule{0.2em}{0ex}}y\phantom{\rule{0.2em}{0ex}}\text{on both sides.}\hfill \\ \hfill \frac{dy}{dx}& =\hfill & {x}^{r}\frac{r}{x}\hfill & & & \text{Step 5. Substitute}\phantom{\rule{0.2em}{0ex}}y={x}^{r}.\hfill \\ \hfill \frac{dy}{dx}& =\hfill & r{x}^{r-1}\hfill & & & \text{Simplify.}\hfill \end{array}

Use logarithmic differentiation to find the derivative of y=xx.y={x}^{x}.

dydx=xx(1+lnx)\frac{dy}{dx}={x}^{x}\left(1+\text{ln}\phantom{\rule{0.1em}{0ex}}x\right)

Hint

Follow the problem solving strategy.

Find the derivative of y=(tanx)π.y={\left(\text{tan}\phantom{\rule{0.1em}{0ex}}x\right)}^{\pi }.

y′=π(tanx)π−1sec2x{y}^{\prime }=\pi {\left(\text{tan}\phantom{\rule{0.1em}{0ex}}x\right)}^{\pi -1}{\text{sec}}^{2}x

Hint

Use the result from [link].

Key Concepts

  • On the basis of the assumption that the exponential function y=bx,b>0y={b}^{x},b>0 is continuous everywhere and differentiable at 0, this function is differentiable everywhere and there is a formula for its derivative.
  • We can use a formula to find the derivative of y=lnx,y=\text{ln}\phantom{\rule{0.1em}{0ex}}x, and the relationship logbx=lnxlnb{\text{log}}_{b}x=\frac{\text{ln}\phantom{\rule{0.1em}{0ex}}x}{\text{ln}\phantom{\rule{0.1em}{0ex}}b} allows us to extend our differentiation formulas to include logarithms with arbitrary bases.
  • Logarithmic differentiation allows us to differentiate functions of the form y=g(x)f(x)y=g{\left(x\right)}^{f\left(x\right)} or very complex functions by taking the natural logarithm of both sides and exploiting the properties of logarithms before differentiating.

Key Equations

  • Derivative of the natural exponential function

    ddx(eg(x))=eg(x)g′(x)\frac{d}{dx}\left({e}^{g\left(x\right)}\right)={e}^{g\left(x\right)}{g}^{\prime }\left(x\right)
  • Derivative of the natural logarithmic function

    ddx(lng(x))=1g(x)g′(x)\frac{d}{dx}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}g\left(x\right)\right)=\frac{1}{g\left(x\right)}{g}^{\prime }\left(x\right)
  • Derivative of the general exponential function

    ddx(bg(x))=bg(x)g′(x)lnb\frac{d}{dx}\left({b}^{g\left(x\right)}\right)={b}^{g\left(x\right)}{g}^{\prime }\left(x\right)\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}b
  • Derivative of the general logarithmic function

    ddx(logbg(x))=g′(x)g(x)lnb\frac{d}{dx}\left({\text{log}}_{b}g\left(x\right)\right)=\frac{{g}^{\prime }\left(x\right)}{g\left(x\right)\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}b}

For the following exercises, find f′(x){f}^{\prime }\left(x\right) for each function.

f(x)=x2exf\left(x\right)={x}^{2}{e}^{x}

2xex+x2ex2x{e}^{x}+{x}^{2}{e}^{x}

f(x)=e−xxf\left(x\right)=\frac{{e}^{\text{−}x}}{x}

f(x)=ex3lnxf\left(x\right)={e}^{{x}^{3}\text{ln}\phantom{\rule{0.1em}{0ex}}x}

ex3lnx(3x2lnx+x2){e}^{{x}^{3}\text{ln}\phantom{\rule{0.1em}{0ex}}x}\left(3{x}^{2}\text{ln}\phantom{\rule{0.1em}{0ex}}x+{x}^{2}\right)

f(x)=e2x+2xf\left(x\right)=\sqrt{{e}^{2x}+2x}

f(x)=ex−e−xex+e−xf\left(x\right)=\frac{{e}^{x}-{e}^{\text{−}x}}{{e}^{x}+{e}^{\text{−}x}}

4(ex+e−x)2\frac{4}{{\left({e}^{x}+{e}^{\text{−}x}\right)}^{2}}

f(x)=10xln10f\left(x\right)=\frac{{10}^{x}}{\text{ln}\phantom{\rule{0.1em}{0ex}}10}

f(x)=24x+4x2f\left(x\right)={2}^{4x}+4{x}^{2}

24x+2⋅ln2+8x{2}^{4x+2}·\text{ln}\phantom{\rule{0.1em}{0ex}}2+8x

f(x)=3sin3xf\left(x\right)={3}^{\text{sin}\phantom{\rule{0.1em}{0ex}}3\text{x}}

f(x)=xπ⋅πxf\left(x\right)={x}^{\pi }·{\pi }^{x}

πxπ−1⋅πx+xπ⋅πxlnπ\pi {x}^{\pi -1}·{\pi }^{x}+{x}^{\pi }·{\pi }^{x}\text{ln}\phantom{\rule{0.1em}{0ex}}\pi

f(x)=ln(4x3+x)f\left(x\right)=\text{ln}\phantom{\rule{0.1em}{0ex}}\left(4{x}^{3}+x\right)

f(x)=ln5x−7f\left(x\right)=\text{ln}\sqrt{5x-7}

52(5x−7)\frac{5}{2\left(5x-7\right)}

f(x)=x2ln9xf\left(x\right)={x}^{2}\text{ln}\phantom{\rule{0.1em}{0ex}}9x

f(x)=log(secx)f\left(x\right)=\text{log}\phantom{\rule{0.1em}{0ex}}\left(\text{sec}\phantom{\rule{0.1em}{0ex}}x\right)

tanxln10\frac{\text{tan}\phantom{\rule{0.1em}{0ex}}x}{\text{ln}\phantom{\rule{0.1em}{0ex}}10}

f(x)=log7(6x4+3)5f\left(x\right)={\text{log}}_{7}{\left(6{x}^{4}+3\right)}^{5}

f(x)=2x⋅log37x2−4f\left(x\right)={2}^{x}·{\text{log}}_{3}{7}^{{x}^{2}-4}

2x⋅ln2⋅log37x2−4+2x⋅2xln7ln3{2}^{x}·\text{ln}\phantom{\rule{0.1em}{0ex}}2·{\text{log}}_{3}{7}^{{x}^{2}-4}+{2}^{x}·\frac{2x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}7}{\text{ln}\phantom{\rule{0.1em}{0ex}}3}

For the following exercises, use logarithmic differentiation to find dydx.\frac{dy}{dx}.

y=xxy={x}^{\sqrt{x}}

y=(sin2x)4xy={\left(\text{sin}\phantom{\rule{0.1em}{0ex}}2x\right)}^{4x}

(sin2x)4x[4⋅ln(sin2x)+8x⋅cot2x]{\left(\text{sin}\phantom{\rule{0.1em}{0ex}}2x\right)}^{4x}\left[4·\text{ln}\phantom{\rule{0.1em}{0ex}}\left(\text{sin}\phantom{\rule{0.1em}{0ex}}2x\right)+8x·\text{cot}\phantom{\rule{0.1em}{0ex}}2x\right]

y=(lnx)lnxy={\left(\text{ln}\phantom{\rule{0.1em}{0ex}}x\right)}^{\text{ln}\phantom{\rule{0.1em}{0ex}}x}

y=xlog2xy={x}^{{\text{log}}_{2}x}

xlog2x⋅2lnxxln2{x}^{{\text{log}}_{2}x}·\frac{2\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}x}{x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}2}

y=(x2−1)lnxy={\left({x}^{2}-1\right)}^{\text{ln}\phantom{\rule{0.1em}{0ex}}x}

y=xcotxy={x}^{\text{cot}\phantom{\rule{0.1em}{0ex}}x}

xcotx⋅[−csc2x⋅lnx+cotxx]{x}^{\text{cot}\phantom{\rule{0.1em}{0ex}}x}·\left[\text{−}{\text{csc}}^{2}x·\text{ln}\phantom{\rule{0.1em}{0ex}}x+\frac{\text{cot}\phantom{\rule{0.1em}{0ex}}x}{x}\right]

y=x+11x2−43y=\frac{x+11}{\sqrt[3]{{x}^{2}-4}}

y=x−1/2(x2+3)2/3(3x−4)4y={x}^{-1\text{/}2}{\left({x}^{2}+3\right)}^{2\text{/}3}{\left(3x-4\right)}^{4}

x−1/2(x2+3)2/3(3x−4)4⋅[−12x+4x3(x2+3)+123x−4]{x}^{-1\text{/}2}{\left({x}^{2}+3\right)}^{2\text{/}3}{\left(3x-4\right)}^{4}·\left[\frac{-1}{2x}+\frac{4x}{3\left({x}^{2}+3\right)}+\frac{12}{3x-4}\right]

[T] Find an equation of the tangent line to the graph of f(x)=4xe(x2−1)f\left(x\right)=4x{e}^{\left({x}^{2}-1\right)} at the point where

x=−1.x=-1. Graph both the function and the tangent line.

[T] Find the equation of the line that is normal to the graph of f(x)=x⋅5xf\left(x\right)=x·{5}^{x} at the point where x=1.x=1. Graph both the function and the normal line.


The function starts at (−3, 0), decreases slightly and then increases through the origin and increases to (1.25, 10). There is a straight line marked T(x) with slope −1/(5 + 5 ln 5) and y intercept 5 + 1/(5 + 5 ln 5).

y=−15+5ln5x+(5+15+5ln5)y=\frac{-1}{5+5\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}5}x+\left(5+\frac{1}{5+5\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}5}\right)

[T] Find the equation of the tangent line to the graph of x3−xlny+y3=2x+5{x}^{3}-x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}y+{y}^{3}=2x+5 at the point where x=2.x=2. (Hint: Use implicit differentiation to find \frac{dy}{dx}.\right) Graph both the curve and the tangent line.

Consider the function y=x1/xy={x}^{1\text{/}x} for x>0.x>0.

  1. Determine the points on the graph where the tangent line is horizontal.
  2. Determine the points on the graph where y′>0{y}^{\prime }>0 and those where y′<0.{y}^{\prime }<0.

a. x=e 2.718x=e~2.718 b. (e,∞),(0,e)\left(e,\infty \right),\left(0,e\right)

The formula I(t)=sintetI\left(t\right)=\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}t}{{e}^{t}} is the formula for a decaying alternating current.

  1. Complete the following table with the appropriate values.

    ttsintet\frac{\text{sin}\phantom{\rule{0.1em}{0ex}}t}{{e}^{t}}
    0(i)
    π2\frac{\pi }{2}(ii)
    π\pi (iii)
    3π2\frac{3\pi }{2}(iv)
    2π2\pi (v)
    2π2\pi (vi)
    3π3\pi (vii)
    7π2\frac{7\pi }{2}(viii)
    4π4\pi (ix)
  2. Using only the values in the table, determine where the tangent line to the graph of I(t)I\left(t\right) is horizontal.

[T] The population of Toledo, Ohio, in 2000 was approximately 500,000. Assume the population is increasing at a rate of 5% per year.

  1. Write the exponential function that relates the total population as a function of t.t.
  2. Use a. to determine the rate at which the population is increasing in tt years.
  3. Use b. to determine the rate at which the population is increasing in 10 years.

a. P=500,000(1.05)tP=500,000{\left(1.05\right)}^{t} individuals b. P′(t)=24395⋅(1.05)t{P}^{\prime }\left(t\right)=24395·{\left(1.05\right)}^{t} individuals per year c. 39,73739,737 individuals per year

[T] An isotope of the element erbium has a half-life of approximately 12 hours. Initially there are 9 grams of the isotope present.

  1. Write the exponential function that relates the amount of substance remaining as a function of t,t, measured in hours.
  2. Use a. to determine the rate at which the substance is decaying in tt hours.
  3. Use b. to determine the rate of decay at t=4t=4 hours.

[T] The number of cases of influenza in New York City from the beginning of 1960 to the beginning of 1961 is modeled by the function

N(t)=5.3e0.093t2−0.87t,(0≤t≤4),N\left(t\right)=5.3{e}^{0.093{t}^{2}-0.87t},\left(0\le t\le 4\right),

where N(t)N\left(t\right) gives the number of cases (in thousands) and t is measured in years, with t=0t=0 corresponding to the beginning of 1960.

  1. Show work that evaluates N(0)N\left(0\right) and N(4).N\left(4\right). Briefly describe what these values indicate about the disease in New York City.
  2. Show work that evaluates N′(0){N}^{\prime }\left(0\right) and N′(3).{N}^{\prime }\left(3\right). Briefly describe what these values indicate about the disease in the United States.

a. At the beginning of 1960 there were 5.3 thousand cases of the disease in New York City. At the beginning of 1963 there were approximately 723 cases of the disease in the United States. b. At the beginning of 1960 the number of cases of the disease was decreasing at rate of −4.611-4.611 thousand per year; at the beginning of 1963, the number of cases of the disease was decreasing at a rate of −0.2808-0.2808 thousand per year.

[T] The relative rate of change of a differentiable function y=f(x)y=f\left(x\right) is given by \frac{100·{f}^{\prime }\left(x\right)}{f\left(x\right)}\text{%}. One model for population growth is a Gompertz growth function, given by P(x)=ae−b⋅e−cxP\left(x\right)=a{e}^{\text{−}b·{e}^{\text{−}cx}} where a,b,a,b, and cc are constants.

  1. Find the relative rate of change formula for the generic Gompertz function.
  2. Use a. to find the relative rate of change of a population in x=20x=20 months when a=204,b=0.0198,a=204,b=0.0198, and c=0.15.c=0.15.
  3. Briefly interpret what the result of b. means.

For the following exercises, use the population of New York City from 1790 to 1860, given in the following table.

New York City Population Over TimeSource: http://en.wikipedia.org/wiki/Largest_cities_in_the_United_States
_by_population_by_decade.Years since 1790Population033,1311060,5152096,37330123,70640202,30050312,71060515,54770813,669

[T] Using a computer program or a calculator, fit a growth curve to the data of the form p=abt.p=a{b}^{t}.

p=35741(1.045)tp=35741{\left(1.045\right)}^{t}

[T] Using the exponential best fit for the data, write a table containing the derivatives evaluated at each year.

[T] Using the exponential best fit for the data, write a table containing the second derivatives evaluated at each year.

Years since 1790P″P\text{″}
069.25
10107.5
20167.0
30259.4
40402.8
50625.5
60971.4
701508.5

[T] Using the tables of first and second derivatives and the best fit, answer the following questions:

  1. Will the model be accurate in predicting the future population of New York City? Why or why not?
  2. Estimate the population in 2010. Was the prediction correct from a.?

Chapter Review Exercises

True or False? Justify the answer with a proof or a counterexample.

Every function has a derivative.

False.

A continuous function has a continuous derivative.

A continuous function has a derivative.

False

If a function is differentiable, it is continuous.

Use the limit definition of the derivative to exactly evaluate the derivative.

f(x)=x+4f\left(x\right)=\sqrt{x+4}

12x+4\frac{1}{2\sqrt{x+4}}

f(x)=3xf\left(x\right)=\frac{3}{x}

Find the derivatives of the following functions.

f(x)=3x3−4x2f\left(x\right)=3{x}^{3}-\frac{4}{{x}^{2}}

9x2+8x39{x}^{2}+\frac{8}{{x}^{3}}

f(x)=(4−x2)3f\left(x\right)={\left(4-{x}^{2}\right)}^{3}

f(x)=esinxf\left(x\right)={e}^{\text{sin}\phantom{\rule{0.1em}{0ex}}x}

esinxcosx{e}^{\text{sin}\phantom{\rule{0.1em}{0ex}}x}\text{cos}\phantom{\rule{0.1em}{0ex}}x

f(x)=ln(x+2)f\left(x\right)=\text{ln}\phantom{\rule{0.1em}{0ex}}\left(x+2\right)

f(x)=x2cosx+xtan(x)f\left(x\right)={x}^{2}\text{cos}\phantom{\rule{0.1em}{0ex}}x+x\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}\left(x\right)

xsec2(x)+2xcos(x)+tan(x)−x2sin(x)x\phantom{\rule{0.1em}{0ex}}{\text{sec}}^{2}\left(x\right)+2x\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}\left(x\right)+\text{tan}\phantom{\rule{0.1em}{0ex}}\left(x\right)-{x}^{2}\text{sin}\phantom{\rule{0.1em}{0ex}}\left(x\right)

f(x)=3x2+2f\left(x\right)=\sqrt{3{x}^{2}+2}

f(x)=x4sin−1(x)f\left(x\right)=\frac{x}{4}\phantom{\rule{0.1em}{0ex}}{\text{sin}}^{-1}\left(x\right)

14(x1−x2+sin−1(x))\frac{1}{4}\left(\frac{x}{\sqrt{1-{x}^{2}}}+{\text{sin}}^{-1}\left(x\right)\right)

x2y=(y+2)+xysin(x){x}^{2}y=\left(y+2\right)+xy\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}\left(x\right)

Find the following derivatives of various orders.

First derivative of y=xln(x)cosxy=x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}\left(x\right)\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}x

cosx⋅(lnx+1)−xln(x)sinx\text{cos}\phantom{\rule{0.1em}{0ex}}x·\left(\text{ln}\phantom{\rule{0.1em}{0ex}}x+1\right)-x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}\left(x\right)\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}x

Third derivative of y=(3x+2)2y={\left(3x+2\right)}^{2}

Second derivative of y=4x+x2sin(x)y={4}^{x}+{x}^{2}\text{sin}\phantom{\rule{0.1em}{0ex}}\left(x\right)

4x(ln4)2+2sinx+4xcosx−x2sinx{4}^{x}{\left(\text{ln}\phantom{\rule{0.1em}{0ex}}4\right)}^{2}+2\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}x+4x\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}x-{x}^{2}\text{sin}\phantom{\rule{0.1em}{0ex}}x

Find the equation of the tangent line to the following equations at the specified point.

y=cos−1(x)+xy={\text{cos}}^{-1}\left(x\right)+x at x=0x=0

y=x+ex−1xy=x+{e}^{x}-\frac{1}{x} at x=1x=1

T=(2+e)x−2T=\left(2+e\right)x-2

Draw the derivative for the following graphs.

The function begins at (−3, 0.5) and decreases to a local minimum at (−2.3, −2). Then the function increases through (−1.5, 0) and slows its increase through (0, 2). It then slowly increases to a local maximum at (2.3, 6) before decreasing to (3, 3).
The function decreases linearly from (−1, 4) to the origin, at which point it increases as x2, passing through (1, 1) and (2, 4).


The function is the straight line y = −4 until x = 0, at which point it becomes a straight line starting at the origin with slope 2. There is no value assigned for this function at x = 0.

The following questions concern the water level in Ocean City, New Jersey, in January, which can be approximated by w(t)=1.9+2.9cos(π6t),w\left(t\right)=1.9+2.9\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}\left(\frac{\pi }{6}t\right), where t is measured in hours after midnight, and the height is measured in feet.

Find and graph the derivative. What is the physical meaning?

Find w′(3).{w}^{\prime }\left(3\right). What is the physical meaning of this value?

w′(3)=−2.9π6.{w}^{\prime }\left(3\right)=-\frac{2.9\pi }{6}. At 3 a.m. the tide is decreasing at a rate of 1.514 ft/hr.

The following questions consider the wind speeds of Hurricane Katrina, which affected New Orleans, Louisiana, in August 2005. The data are displayed in a table.

Wind Speeds of Hurricane KatrinaSource: http://news.nationalgeographic.com/news/2005/09/0914_050914_katrina_timeline.html.
Hours after Midnight, August 26Wind Speed (mph)
145
575
11100
29115
49145
58175
73155
81125
8595
10735

Using the table, estimate the derivative of the wind speed at hour 39. What is the physical meaning?

Estimate the derivative of the wind speed at hour 83. What is the physical meaning?

−7.5.-7.5. The wind speed is decreasing at a rate of 7.5 mph/hr

Glossary

logarithmic differentiation
is a technique that allows us to differentiate a function by first taking the natural logarithm of both sides of an equation, applying properties of logarithms to simplify the equation, and differentiating implicitly