We can figure it out

$\newenvironment {prompt}{}{} \newcommand {\accordion }[0]{} \newcommand {\ungraded }[0]{} \newcommand {\xmDefaultPreamble }[0]{xmPreamble.tex} \newcommand {\xmDefaultPrintstyle }[0]{xmPrintstyle.sty} \newcommand {\todo }[0]{} \newcommand {\oiint }[0]{{\large \bigcirc }\kern -1.56em\iint } \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{}$

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.

2025-01-06 19:56:35