# 3.9 Derivatives of exponential and logarithmic functions  (Page 3/6)

 Page 3 / 6

## Proof

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

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

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

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

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

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

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

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

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

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

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

## Taking a derivative of a natural logarithm

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

$\begin{array}{ccccc}\hfill {f}^{\prime }\left(x\right)& =\frac{1}{{x}^{3}+3x-4}·\left(3{x}^{2}+3\right)\hfill & & & \text{Use}\phantom{\rule{0.2em}{0ex}}g\left(x\right)={x}^{3}+3x-4\phantom{\rule{0.2em}{0ex}}\text{in}\phantom{\rule{0.2em}{0ex}}{h}^{\prime }\left(x\right)=\frac{1}{g\left(x\right)}{g}^{\prime }\left(x\right).\hfill \\ & =\frac{3{x}^{2}+3}{{x}^{3}+3x-4}\hfill & & & \text{Rewrite.}\hfill \end{array}$

## Using properties of logarithms in a derivative

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

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.

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

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

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

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

## Derivatives of general exponential and logarithmic functions

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

1. If, $y={\text{log}}_{b}x,$ then
$\frac{dy}{dx}=\frac{1}{x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}b}.$

More generally, if $h\left(x\right)={\text{log}}_{b}\left(g\left(x\right)\right),$ then for all values of x for which $g\left(x\right)>0,$
${h}^{\prime }\left(x\right)=\frac{{g}^{\prime }\left(x\right)}{g\left(x\right)\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}b}.$
2. If $y={b}^{x},$ then
$\frac{dy}{dx}={b}^{x}\text{ln}\phantom{\rule{0.1em}{0ex}}b.$

More generally, if $h\left(x\right)={b}^{g\left(x\right)},$ then
${h}^{\prime }\left(x\right)={b}^{g\left(x\right)}g\text{″}\left(x\right)\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}b.$

## Proof

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

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

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

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

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

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

## Applying derivative formulas

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

Use the quotient rule and [link] .

$\begin{array}{ccccc}\hfill {h}^{\prime }\left(x\right)& =\frac{{3}^{x}\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}3\left({3}^{x}+2\right)-{3}^{x}\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}3\left({3}^{x}\right)}{{\left({3}^{x}+2\right)}^{2}}\hfill & & & \text{Apply the quotient rule.}\hfill \\ & =\frac{2·{3}^{x}\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}3}{{\left({3}^{x}+2\right)}^{2}}\hfill & & & \text{Simplify.}\hfill \end{array}$

## Finding the slope of a tangent line

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

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

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

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

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

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

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

## Logarithmic differentiation

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

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