We can figure it out

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\todo }[0]{} \newcommand {\oiint }[0]{{\large \bigcirc }\kern -1.56em\iint } \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\d }[0]{\,d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{\bigg [ #1 \bigg ]} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\phi }[0]{\varphi } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\scal }{scal} \DeclareMathOperator {\sign }{sign} \newcommand {\arrowvec }[1]{{\overset {\rightharpoonup }{#1}}} \newcommand {\vec }[1]{{\overset {\boldsymbol {\rightharpoonup }}{\mathbf {#1}}}\hspace {0in}} \newcommand {\point }[1]{\left (#1\right )} \newcommand {\pt }[1]{\mathbf {#1}} \newcommand {\Lim }[2]{\lim _{\point {#1} \to \point {#2}}} \DeclareMathOperator {\proj }{\mathbf {proj}} \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\vec {\boldsymbol {\l }}} \newcommand {\uvec }[1]{\mathbf {\hat {#1}}} \newcommand {\utan }[0]{\mathbf {\hat {t}}} \newcommand {\unormal }[0]{\mathbf {\hat {n}}} \newcommand {\ubinormal }[0]{\mathbf {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\lto }[0]{\mathop {\longrightarrow \,}\limits } \newcommand {\bar }[0]{\overline } \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument } \newcommand {\descriptionlabel }[1]{\hspace {\labelsep }\textbf {#1:}}$

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 $\begin{align*} \ddx \tan^{-1}(x) &= -\tan^{-2}(x)\sec^2(x)\\ &= -\frac{\cos^2x}{\sin^2x}\cdot \frac{1}{\cos^2x}\\ &= \frac{-1}{\sin^2x}\\ &= -\csc^2x \end{align*}$
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 $\ddx \arctan (x) = -\csc ^2(x)?$

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.