Inverse functions and inverse trigonometry

Understand the idea

An inverse derivative uses the slope of the original function at the corresponding input, not at the same numeric argument. Require local invertibility and a nonzero original derivative. The derivatives of inverse trigonometric functions can be derived implicitly; apply the chain rule to their arguments. The principal branches determine the relevant domains.

(f^{-1})'(b)=\frac1{f'(a)}\quad\text{where }f(a)=b;\qquad(\arcsin x)'=\frac1{\sqrt{1-x^2}},\quad(\arccos x)'=-\frac1{\sqrt{1-x^2}},\quad(\arctan x)'=\frac1{1+x^2}

Worked example

If $\displaystyle f(2)=5$ and $\displaystyle f'(2)=3$ , find $\displaystyle (f^{-1})'(5)$ .

The inverse sends

\displaystyle 5

to

\displaystyle 2

. Reciprocal slopes at corresponding points give

\displaystyle (f^{-1})'(5)=1/f'(2)=1/3

.

Try it yourself

Differentiate $\displaystyle y=\arctan(3x)$ .

Show solution and reasoning

The chain rule gives $\displaystyle y'=3/(1+9x^2)$ .