OBJECTIVES

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

1. Differentiate logarithmic functions (p. 222)
2. Use logarithmic differentiation (p. 225)
3. Express $e$ as a limit (p. 227)

NEED TO REVIEW?

Logarithmic functions are discussed in Section P.5, pp. 42-45.

THEOREM Derivative of a Logarithmic Function

The derivative of the logarithmic function $y=\log _{a}x$, $x>0$, $a>0$, and $a≠ 1$, is $$\begin{equation*} \bbox[5px, border:1px solid black, #F9F7ED]{ y^\prime =\dfrac{d}{dx}\log _{a}x=\dfrac{1}{x\ln a}\qquad x > 0}\tag{1} \end{equation*}$$

THEOREM Derivative of the Natural Logarithm Function

The derivative of the natural logarithm function $y=\ln x,$ $x>0,$ is $$\bbox[5px, border:1px solid black, #F9F7ED]{ y^\prime =\dfrac{d}{dx}\ln x=\dfrac{1}{x}}$$

Find $y^\prime$ if:

Solution (a) We use the Power Rule for Functions. Then $$\begin{eqnarray*} && y^\prime =\dfrac{d}{dx}( \ln x) ^{2} \underset{\underset{\color{#0066A7}{\hbox{\(\tfrac{d}{dx}[u(x) ] ^{n}{=n}[u(x) ]^{n-1}\tfrac{du}{dx}\)}}}{\color{#0066A7}{\uparrow}}}{=} 2\ln x\cdot \dfrac{d}{dx}\ln x= (2\ln x)\left(\dfrac{1}{x}\right)=\dfrac{2\ln x}{x}\\ \end{eqnarray*}$$

(b) Remember that $\log x=\log _{10}x$. Then $y=\dfrac{1}{2}\log x=\dfrac{1}{2}\log_{10} x$. $$\begin{eqnarray*} && y^\prime =\dfrac{1}{2}\left( \dfrac{d}{dx}\log_{10} x\right) = \dfrac{1}{2}\cdot \dfrac{1}{x\ln 10}=\dfrac{1}{2x\ln 10}\qquad {\color{#0066A7}{\tfrac{d}{dx}\log_{a}{x} = \tfrac{1}{x\ln a}}} \end{eqnarray*}$$

(c) We assume $y$ is a differentiable function of $x$ and use implicit differentiation. Then $$\begin{eqnarray*} \begin{array}{rl@{\quad}l} \dfrac{d}{dx}\left( \ln x+\ln y\right) &= \dfrac{d}{dx}(2x) \\ \dfrac{d}{dx}\ln x+\dfrac{d}{dx}\ln y &= 2 \\ \dfrac{1}{x}+\dfrac{1}{y}\dfrac{dy}{dx} &= 2 & {\color{#0066A7}{\tfrac{d}{dx}\ln {y}=\tfrac{1}{y}\tfrac{dy}{dx}}}\\ y^\prime &= \dfrac{dy}{dx}=y\left(2-\dfrac{1}{x}\right) =\dfrac{y\left(2x-1\right)}{x} & {\color{#0066A7}{\hbox{Solve for}\quad {y^\prime }=\tfrac{dy}{dx}.}} \end{array} \end{eqnarray*}$$

NOW WORK

Problem 9.

Find $y^\prime$ if:

Solution

(a) $$\begin{eqnarray*} && y^\prime = \dfrac{d}{dx}\ln (5x) = \dfrac{\dfrac{d}{dx}(5x) }{5x}=\dfrac{5}{5x} = \dfrac{1}{x} \qquad {\color{#0066A7}{\tfrac{d}{dx}\ln u({x}) = \tfrac{u^\prime (x)}{u(x)}}} \end{eqnarray*}$$

(b) $$\begin{eqnarray*} y^\prime = \dfrac{d}{dx}[ x\ln (x^{2}+1)] &=& x\dfrac{d}{dx} \ln (x^{2}+1) +\left( \dfrac{d}{dx}x\right) \ln (x^{2}+1)\qquad{\color{#0066A7}{\hbox{Product Rule}}} \\ &&\underset{\underset{\color{#0066A7}{\tfrac{d}{dx}\ln u(x)=\tfrac{u^\prime (x)}{u(x)}}}{\color{#0066A7}{\uparrow}}}{=} x\cdot \dfrac{\dfrac{d}{dx}(x^{2}+1) }{ x^{2}+1}+1\cdot \ln (x^{2}+1) =\dfrac{2x^{2}}{x^{2}+1}+\ln (x^{2}+1) \end{eqnarray*}$$

(c) $$\begin{eqnarray*} y &=&\ln \vert x\vert =\left\{ \begin{array}{c@{\quad}l} \ln x & \hbox{if}\; x>0 \\ \ln (-x) & \hbox{if}\; x<0 \end{array} \right. \\ y^\prime &=& \dfrac{d}{dx}\ln \vert x\vert =\left\{ \begin{array}{c@{\quad}l} \dfrac{1}{x} & \hbox{if}\;x>0 \\ \dfrac{1}{-x}(-1) =\dfrac{1}{x} & \hbox{if}\;x<0 \end{array}\right. \end{eqnarray*}$$

That is, $\dfrac{d}{dx}\ln \vert x\vert =\dfrac{1}{x}$ for all numbers $x≠ 0$.

NOW WORK

Problem 17.

NEED TO REVIEW?

Properties of logarithms are discussed in Appendix A.1, pp. A-10 to A-11.

Find $y^\prime$ if $y=\ln \left[ \dfrac{\left( 2x-1\right) ^{3}\sqrt{2x^{4}+1}}{x}\right]\!$.

Solution Rather than attempting to use the Chain Rule, Quotient Rule, and Product Rule, we first simplify the right side by using properties of logarithms. $$\begin{eqnarray*} y &=& \ln \left[ \dfrac{\left( 2x-1\right) ^{3}\sqrt{2x^{4}+1}}{x}\right] =\ln \left( 2x-1\right) ^{3}+\ln \sqrt{2x^{4}+1}-\ln x\\ &=& 3\ln \left( 2x-1\right) + \dfrac{1}{2}\ln ( 2x^{4}+1) -\ln x \end{eqnarray*}$$

Now we differentiate $y.$ $$\begin{eqnarray*} y^\prime &=&\dfrac{d}{dx}\left[ 3\ln \left( 2x-1\right) +\dfrac{1}{2}\ln ( 2x^{4}+1) -\ln x\right]\\ &=&\dfrac{d}{dx}\left[3\ln \left( 2x-1\right)\right] + \dfrac{d}{dx}\left[\dfrac{1}{2}\ln ( 2x^{4}+1)\right] -\dfrac{d}{dx}\ln x \\ &=&3\cdot \dfrac{2}{2x-1}+\dfrac{1}{2}\cdot \dfrac{8x^{3}}{2x^{4}+1}-\dfrac{1}{x}=\dfrac{6}{2x-1}+\dfrac{4x^{3}}{2x^{4}+1}-\dfrac{1}{x} \end{eqnarray*}$$

NOW WORK

Problem 27.

Find $y^\prime$ if $y=\dfrac{x^{2}\sqrt{5x+1}}{( 3x-2) ^{3}}$.

Solution It is easier to find $y^\prime$ if we take the natural logarithm of each side before differentiating. That is, we write $$\ln y=\ln \left[ \dfrac{x^{2}\sqrt{5x+1}}{( 3x-2) ^{3}}\right]$$

and simplify the equation using properties of logarithms. $$\begin{eqnarray*} \ln y &=&\ln [ x^{2}\sqrt{5x+1}] -\ln ( 3x-2) ^{3}=\ln x^{2}+\ln (5x+1) ^{1/2}-\ln ( 3x-2) ^{3} \\ &=&2\ln x+\dfrac{1}{2}\ln (5x+1) -3\ln ( 3x-2) \end{eqnarray*}$$

To find $y^\prime$, we use implicit differentiation. $$\begin{eqnarray*} \dfrac{d}{dx}\ln y&=&\dfrac{d}{dx}\left[ 2\ln x+\dfrac{1}{2}\ln (5x+1) -3\ln (3x-2) \right]\\ \dfrac{y^\prime}{y}&=&\dfrac{d}{dx}(2\ln x) + \dfrac{d}{dx} \left[\dfrac{1}{2}\ln (5x+1)\right]-\dfrac{d}{dx}[3\ln(3x-2)]\\ &=&\dfrac{2}{x}+\dfrac{5}{2 ( 5x+1) }-\dfrac{9}{3x-2} \\ y^\prime &=&y\!\left[ \dfrac{2}{x}+\dfrac{5}{2 ( 5x+1) }-\dfrac{9}{3x-2}\right] \!=\!\left[ \dfrac{x^{2}\sqrt{5x+1}}{( 3x-2) ^{3}}\right]\!\! \left[ \dfrac{2}{x}+\dfrac{5}{2(5x+1) }-\dfrac{9}{3x-2}\right] \end{eqnarray*}$$

Steps for Using Logarithmic Differentiation

Step 1 If the function $y=f( x)$ consists of products, quotients, and powers, take the natural logarithm of each side. Then simplify the equation using properties of logarithms.

Step 2 Differentiate implicitly, and use the fact that $\dfrac{d}{dx}\ln y=\dfrac{y^\prime }{y}$.

Step 3 Solve for $y^\prime$, and replace $y$ with $f( x)$.

NOW WORK

Problem 51.

Find $y^\prime$ if $y=x^{x}$, $x>0$.

Solution Notice that $x^{x}$ is neither $x$ raised to a fixed power $a$, nor a fixed base $a$ raised to a variable power. We follow the steps for logarithmic differentiation:

Step 1 Take the natural logarithm of each side of $y=x^{x}$, and simplify: $$\ln y=\ln x^{x}=x\ln x$$

Step 2 Differentiate implicitly. $$\begin{eqnarray*} \dfrac{d}{dx}\ln y&=&\dfrac{d}{dx}(x\ln x)\\ \dfrac{y^\prime }{y} &=& x\cdot \dfrac{d}{dx}\ln x+\left( \dfrac{d}{dx}x\right) \cdot \ln x=x\left(\dfrac{1}{x}\right) +1\cdot \ln x=1+\ln x \end{eqnarray*}$$

Step 3 Solve for $y^\prime$: $y^\prime =y( 1+\ln x) =x^{x}( 1+\ln x).$

NOW WORK

Problem 57.

Find an equation of the tangent line to the graph of $f( x) =\dfrac{x\sqrt{x^{2}+3}}{1+x}$ at the point $(1,1) .$

Solution The slope of the tangent line to the graph of $y=f( x)$ at the point $(1,1)$ is $f^\prime (1)$. Since the function consists of a product, a quotient, and a power, we follow the steps for logarithmic differentiation:

Step 1 Take the natural logarithm of each side, and simplify: $$\ln y=\ln \dfrac{x\sqrt{x^{2}+3}}{1+x}=\ln x+\dfrac{1}{2}\ln ( x^{2}+3) -\ln ( 1+x)$$

Step 2 Differentiate implicitly. $$\dfrac{y^\prime }{y}=\dfrac{1}{x}+\dfrac{1}{2}\cdot \dfrac{2x}{x^{2}+3}- \dfrac{1}{1+x}=\dfrac{1}{x}+\dfrac{x}{x^{2}+3}-\dfrac{1}{1+x}$$

Step 3 Solve for $y^\prime$ and simplify: $$y^\prime =f^\prime (x) = y\left( \dfrac{1}{x}+\dfrac{x}{x^{2}+3}-\dfrac{1}{1+x}\right) = \dfrac{x\sqrt{x^{2}+3}}{1+x}\left( \dfrac{1}{x}+\dfrac{x}{x^{2}+3}-\dfrac{1}{1+x}\right)$$

Now we find the slope of the tangent line by evaluating $f^\prime (1)$. $$f^\prime (1) =\dfrac{\sqrt{4}}{2}\left( 1+\dfrac{1}{4}-\dfrac{1}{2}\right) =\dfrac{3}{4}$$

Then an equation of the tangent line to the graph of $f$ at the point $(1,1)$ is $$\begin{eqnarray*} y-1 &=&\dfrac{3}{4}( x-1) \\ y &=&\dfrac{3}{4}x+\dfrac{1}{4} \end{eqnarray*}$$

NOW WORK

Problem 75.

THEOREM Power Rule

If $a$ is a real number, then $$\bbox[5px, border:1px solid black, #F9F7ED]{ \dfrac{d}{dx}x^{a}=ax^{a-1}}$$

Proof

Let $y=x^{a},$ $a$ real. Then $\ln y=\ln x^{a}=a\ln x.$ $$\begin{eqnarray*} \frac{y^\prime }{y} &=&a\frac{1}{x} \\ y^\prime &=&\frac{ay}{x}=\frac{ax^{a}}{x}=ax^{a-1} \end{eqnarray*}$$

Find the derivative of:

Solution

THEOREM The Number $e$ as a Limit

The number $e$ can be expressed by either of the two limits: $$\bbox[5px, border:1px solid black, #F9F7ED]{ \hbox{(a)} \lim\limits_{h\rightarrow 0}(1+h)^{1/h} = e \qquad \hbox{or}\qquad \hbox{(b)} \lim\limits_{n\rightarrow ∞ }\left( 1+\dfrac{1}{n}\right) ^{n}=e}\tag{2}$$

Proof

(a) The derivative of $f(x) =\ln x$ is $f^\prime (x) =\frac{1}{x}$ so $f^\prime (1) =1$. Then $$\begin{eqnarray*} f^\prime (1) \underset{\underset{\color{#0066A7}{\hbox{Definition of a Derivative}}}{\color{#0066A7}{\uparrow}}}{=} \lim\limits_{h\rightarrow 0}\frac{f(1+h) -f(1) }{h} &=&\lim\limits_{h\rightarrow 0}\dfrac{\ln (1+h) -\ln 1}{h} = \lim\limits_{h\rightarrow 0}\frac{\ln (1+h) -0}{h} \\ &=&\lim\limits_{h\rightarrow 0}\left[ \dfrac{1}{h}\ln (1+h) \right] =\lim\limits_{h\rightarrow 0}\ln (1+h) ^{1/h} \end{eqnarray*}$$

Since $y=\ln x$ is continuous and $f^\prime (1) =1$, $$\lim\limits_{h\rightarrow 0}\ln (1+h) ^{1/h}=\ln \left[ \lim\limits_{h\rightarrow 0}(1+h) ^{1/h}\right] =1$$

Since $\ln e=1$, $$\ln \left[ \lim\limits_{h\rightarrow 0}(1+h) ^{1/h}\right] =\ln e$$

Since $y=\ln x$ is a one-to-one function, $$\lim\limits_{h\rightarrow 0}(1+h) ^{1/h}=e$$

(b) The limit derived in (a) is valid when $h\rightarrow 0^{+}$. So if we let $n=\dfrac{1}{h}$, then as $h\rightarrow 0^{+}$, $n= \dfrac{1}{h}\rightarrow \infty$ and $$e=\lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{1}{n}\right) ^{n}$$

NOTE

Correct to nine decimal places, $e= 2.\;718\;281\;828.$

Express each limit in terms of the number $e$:

Solution (a) This limit resembles $\lim\limits_{h\rightarrow 0}(1+h) ^{1/h}$, and with some manipulation, it can be expressed in terms of $\lim\limits_{h\rightarrow 0}(1+h) ^{1/h}$. $$( 1+2h) ^{1/h}= [ ( 1+2h) ^{1/( 2h) } ] ^{2}$$

Now let $k=2h,$ and note that $h\rightarrow 0$ is equivalent to $2h=k\rightarrow 0$. So, $$\begin{eqnarray*} && \lim\limits_{h\rightarrow 0}( 1+2h) ^{1/h}=\lim\limits_{k\rightarrow 0}\left[ \left( 1+k\right) ^{1/k}\right] ^{2}=\left[ \lim\limits_{k\rightarrow 0}\left( 1+k\right) ^{1/k}\right] ^{2} \underset{\underset{\color{#0066A7}{\hbox{\((2)\)}}}{\color{#0066A7}{\uparrow}}}{=} e^{2} \\[-10pt] && \end{eqnarray*}$$

(b) This limit resembles $\lim\limits_{n\rightarrow \infty} \left( 1+\dfrac{1}{n}\right) ^{n}$. We rewrite it as follows: $$\left( 1+\dfrac{3}{n}\right) ^{2n}=\left[ \left( 1+\dfrac{3}{n}\right) ^{2n} \right] ^{3/3}=\left[ \left( 1+\dfrac{3}{n}\right) ^{n/3}\right] ^{6}$$

Let $k=\dfrac{n}{3}$. Since $n\rightarrow \infty$ is equivalent to $\dfrac{n }{3}=k\rightarrow \infty \,$, we find that $$\begin{eqnarray*} && \lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{3}{n}\right) ^{2n}=\lim\limits_{k\rightarrow \infty }\left[ \left( 1+\dfrac{1}{k}\right) ^{k}\right] ^{6}=\left[ \lim\limits_{k\rightarrow \infty }\left( 1+\dfrac{1}{ k}\right) ^{k}\right] ^{6} \underset{\underset{\color{#0066A7}{\hbox{\((2)\)}}}{\color{#0066A7}{\uparrow}}}{=} e^{6} \end{eqnarray*}$$

NOW WORK

Problem 77.