Extra (Optional) Exercises – Short Day

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.

Extra Exercise 1: Slope and Inverses

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

Question a

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 \(1/r\).

Question b

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:

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

Question c

Rewrite the formula from Question b to \((f^{-1})'(x)=1/f'(f^{-1}(x))\) by substituting the variable \(x\) with \(f^{-1}(x)\),

Question d

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

Question e

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

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