Math 130, Spring 2001, Lab 9

Math 130-03, Lab 9

This lab is about inverse functions, including logarithm functions, and their derivatives. Logarithm functions are defined as inverse functions of exponential functions. It's a good idea to be familiar with the exponential functions, which are of the form for some positive number . Move the slider in the following applet to see how the function depends on the value of .

There is one particular exponential function that is very important in calculus. The number , which about 2.718, has the property that $\frac{d}{dx}\big(e^x\big)=e^x$ . That is, the exponential function is its own derivative. It has the property that . To verify this, try entering e^x as the function in the following applet. You can move the tangent line along the graph by clicking and dragging on the graph.

What if you enter a different exponential function, such as or ? Try it and see. Notice that the derivatives of these functions make use of a new function named ``ln''. This function is actually the logarithm function to the base . That is, $\ln(x)=\log_e(x)$ . Thus, $\ln(x)$ is the inverse function of . Try entering $\ln(x)$ as the function in the above applet. Check out the derivative of this function, as reported by the applet.

Exercise 1

If and are inverse functions, then for all in the domain of . We can investigate derivatives of inverse functions using the function composition applet, which you have seen previously. We have already noted that the product of the slope of the tangent lines to the two functions is equal to the slopes of the tangent line to the composition. This is essentially what the chain rule says. Now, we can think about what this says when the two functions are inverse functions. In that case, the composition function is just , and its slope is 1. Click the button to open the applet.

The applet is set up to show an exponential and a logarithmic function. These are inverse functions. You can drag the red square to change the points at which the tangent lines are drawn. Answer the following questions, based on this applet:

(a) As you know from working with the chain rule, you have to be careful about which input values are used for functions in your formulas. In the applet, the tangent line to is shown at the point where the input value is . Explain how the formula $\big(f^{-1}\big)'(f(x)) = 1/f'(x)$ can be deduced from the fact that the slope of the third tangent line in the applet is equal to the product of the slopes of the other two tangent lines. Keep in mind that in this example, $f^{-1}$ is .

(c) Here is an unrelated question about inverse functions. The function tan(x) does not have an inverse, since it is not one-to-one. However, we can restrict the domain of this function to get a one-to-one function. The function f(x)=tan(x), for -pi/2 < x < pi/2, does have an inverse. The inverse function is denoted arctan(x) Use the applet to look at the functions tan(x) and arctan(x). Try to explain what you see.

Exercise 2

The second exercise is to verify the formulas for the derivative of an inverse function. This can be done using the chain rule. The discussion assumes that the inverse of a differentiable function is also differentiable, but that should be easy to believe, based on the graph.

(a) Use the chain rule to show that $\frac{d}{dx}e^{f(x)}=e^{f(x)}\frac{d}{dx}\big(f(x)\big)$ for any differentiable function .

(b) Since $\ln(x)$ is the inverse function of , it satisfies $e^{\ln(x)}=x$ . Apply the $\frac{d}{dx}$ operator to both sides of this equation, and use the result to deduce that $\frac{d}{dx}\ln(x)=\frac{1}{x}$ . Explain your reasoning.

(c) Let be any differentiable function that has an inverse. We know that f^-1(f(x)) = x. Differentiate both sides of this equation and apply the chain rule to show that $\big(f^{-1}\big)'(f(x)) = 1/f'(x)$ .

(d) Let . Note that . Since the derivative of this function is always positive, it has an inverse function. Let be the inverse function. Find the value of . Explain your answer.

Exercise 3

Now that you have the formulas for the derivatives of and $\ln(x)$ , you can combine these formulas with all the other rules that you already know for differentiation. Compute the following derivatives, showing each step in your work:

$\displaystyle \textbf{(a)\ \ }$ $\displaystyle \displaystyle \frac{d}{dx}\,e^x\sin(x)$ $\displaystyle \textbf{(b)\ \ }$ $\displaystyle \displaystyle \frac{d}{dx}\,\sin\big(e^x\big)$

$\displaystyle \noalign{\medskip } \textbf{(c)\ \ }$ $\displaystyle \displaystyle \frac{d}{dx}\,e^{x\sin(x)}$ $\displaystyle \textbf{(d)\ \ }$ $\displaystyle \displaystyle \frac{d}{dx}\,\frac{e^x+e^{-x}}{\ln(x)+1}$

$\displaystyle \noalign{\medskip } \textbf{(e)\ \ }$ $\displaystyle \displaystyle \frac{d}{dx}\,\ln(x^2+1)$ $\displaystyle \textbf{(f)\ \ }$ $\displaystyle \displaystyle \frac{d}{dx}\,\left(e^{2x}\tan\big(e^x\sqrt{x^4+17}\,\big)\right)$

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David Eck, March 2001

$\displaystyle \textbf{(a)\ \ }$	$\displaystyle \displaystyle \frac{d}{dx}\,e^x\sin(x)$	$\displaystyle \textbf{(b)\ \ }$	$\displaystyle \displaystyle \frac{d}{dx}\,\sin\big(e^x\big)$
$\displaystyle \noalign{\medskip } \textbf{(c)\ \ }$	$\displaystyle \displaystyle \frac{d}{dx}\,e^{x\sin(x)}$	$\displaystyle \textbf{(d)\ \ }$	$\displaystyle \displaystyle \frac{d}{dx}\,\frac{e^x+e^{-x}}{\ln(x)+1}$
$\displaystyle \noalign{\medskip } \textbf{(e)\ \ }$	$\displaystyle \displaystyle \frac{d}{dx}\,\ln(x^2+1)$	$\displaystyle \textbf{(f)\ \ }$	$\displaystyle \displaystyle \frac{d}{dx}\,\left(e^{2x}\tan\big(e^x\sqrt{x^4+17}\,\big)\right)$
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