We can figure it out

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Two young mathematicians discuss the derivative of inverse functions.

Check out this dialogue between two calculus students (based on a true story):

Devyn: Riley, I have a calculus question.
Riley: Hit me with it.
Devyn: What’s the derivative of $\arctan (x)$ ?
Riley: Hmmm…we haven’t talked about that yet in our class.
Devyn: I know! But maybe we can figure it out.
Riley: Well, if
$f(x)=e^{x} \text { then } f^{-1}(x)=\ln (x)$ so $\ln (x) = e^{-x}$ and now we can use the chain rule to take its derivative
$\begin{align*} \ddx \ln(x) &= \ddx e^{-x} \\ &= -e^{-x}\\ &= -\ln(x)\\ \end{align*}$
Devyn: But is this right?

Let’s see if we can figure out if Devyn and Riley are correct. Start by looking at a plot of $y = \ln (x)$ :

Let $f(x) = ln(x)$ . Use the plot above to determine the sign of the derivative of $f$ . $f'(x) \begin {prompt} \answer { >} \end {prompt} 0$

On the other hand,

What is the sign of $-e^{-x}$ ? $-e^{-x} \begin {prompt} \answer { <} \end {prompt} 0$

In light of the problems above, is it possible that $\ddx \ln (x) = -e^{-x}?$

yes no

When our friends wrote $\ln (x) =f^{-1}(x)$ wheren $f(x)=e^{x}$ , what do they think the ‘‘ $-1$ ’’ represents? Are they correct?

Riley thinks that we can use the power rule on the $-1$ , which tells us that the students are using $-1$ as an exponent for the exponential function. However, in the case of inverse functions such as $\ln (x)$ , the $-1$ is not an exponent.