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 $\arctan (x) = \tan ^{-1}(x)$ and now we can use the chain rule to take its derivative $\begin{align*} \frac{d}{dx} \tan^{-1}(x) &= -\tan^{-2}(x)\sec^2(x)\\ &= -\frac{\cos^2x}{\sin^2x}\cdot \frac{1}{\cos^2x}\\ &= \frac{-1}{\sin^2x}\\ &= -\csc^2x \end{align*}$
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 $\theta = \arctan (x)$ :

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 $\left (\answer [given]{-\infty },\answer [given]{\infty }\right )$

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 $\frac {d}{dx} \arctan (x) = -\csc ^2(x)?$

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.