
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 and now we can use the chain rule to take its derivative
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 $\theta = \arctan (x)$:

Let $f(x) = \arctan (x)$. Use the plot to determine the intervals(s) where the function $f$ is increasing.

From the graph it seems that the function $f$ is increasing on the interval

On the other hand,

What is the sign of $f'(x)= - \csc ^2 (x)$ on the interval $(0,\pi )$?
positive negative
Complete the sentence below:

Since the sign of $f'(x)= - \csc ^2 (x)$ on the interval $(0,\pi )$ is

positive negative
the function $f$ must be
increasing on the interval $(0, \pi )$ decreasing on the interval $(0, \pi )$

In light of the problems above, is it possible that

yes no
When our friends wrote $\arctan (x) = \tan ^{-1}(x)$, 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 tangent function. However, in the case of inverse functions such as $\arctan (x)$, the $-1$ is not an exponent.