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(section:uge3L)=
# Extra (Optional) Exercises -- Short Day

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These exercises are optional extra material for students who are done with the exercises of the day but have the time for and the interest in more.

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## Extra Exercise 1: Slope and Inverses

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In this exercise you may use the fact that a line located in a plane with slope $r \in \mathbb{R}\setminus \{0\}$ being reflected 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 realise graphically 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 derivative $f'(x)$ in $x$, then the derivative of $f^{-1}$ in $f(x)$ 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 to $(f^{-1})'(x)=1/f'(f^{-1}(x))$ by substituting the variable $x$ with $f^{-1}(x)$,

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### Question d

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

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### Question e

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By definition we have $\cos'(x)=-\sin(x)$ and $\sin'(x)=\cos(x)$, and these formulas may be used freely. Now show that $\tan'(x)=\tan^2(x)+1$, and use this to reach the formula 

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


Remark: A similar approach will give rise to the formulas:

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

and

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