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(section:uge3L-extra)=
# Extra (optional) exercises -- Short Day

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These exercises are optional extra material for students who have finished the exercises of the day and have the time and interest for more.

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## Extra exercise 1: Slope and inverse functions

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In this exercise you may use the fact that reflecting a line in the plane with slope $r \in \mathbb{R}\setminus \{0\}$ about the line $y=x$ gives a line with slope $1/r$.

We consider an invertible function $f: \mathbb R \to \mathbb R$.

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### Question a

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Use the result described in the answer to Exercise 4 from Short Day in Week 2 to see geometrically that if the graph of $f$ has a tangent line through $(x,f(x))$ with slope $r \in \mathbb{R}\setminus \{0\}$, then the graph of $f^{-1}$ has a tangent line through $(f(x),x)$ with slope $1/r$.

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### Question b

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Conclude that if $f$ has a (non-zero) derivative $f'(x)$ at $x$, then the derivative of $f^{-1}$ at $f(x)$ equals $1/f'(x)$. In other words: 

$$(f^{-1})'(f(x))=1/f'(x).$$

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### Question c

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Rewrite the formula from Question b as $(f^{-1})'(x)=1/f'(f^{-1}(x))$ by replacing $x$ with $f^{-1}(x)$.

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### Question d

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Now use the formula from Question c to show that $\mathrm{arctan}'(x)=1/\tan'(\mathrm{arctan}(x))$. 

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### Question e

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You may use without proof that $\cos'(x)=-\sin(x)$ and $\sin'(x)=\cos(x)$. Now show that $\tan'(x)=\tan^2(x)+1$, and use this to derive the formula 

$$\mathrm{arctan}'(x)=1/(x^2+1).$$


Remark: a similar approach gives the formulas

$$\mathrm{arccos}'(x)=-1/\sqrt{1-x^2}$$

and

$$\mathrm{arcsin}'(x)=1/\sqrt{1-x^2}.$$