Extra (optional) exercises – Short Day

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

Extra exercise 1: Slope and inverse functions

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 \(f: \mathbb R \to \mathbb R\).

Question a

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 \(1/r\).

Question b

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:

\[(f^{-1})'(f(x))=1/f'(x).\]

Question c

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

Question d

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

Question e

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

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