Derivatives of Inverse Functions and Inverse Trig

$$f(x)=e^x,\qquad f^{-1}(x)=\ln x$$

The inverse is defined for x>0.

$$(f^{-1})'(x)=\frac1{e^{\ln x}}=\frac1x$$

Evaluate f′ at ln x before taking the reciprocal.

$$f(x)=x^3,\qquad f^{-1}(x)=x^{1/3}$$

The cube function is one-to-one on all real numbers.

$$(f^{-1})'(x)=\frac1{3(x^{1/3})^2}=\frac1{3x^{2/3}}\quad(x\ne0)$$

At x=0 the cube root has a vertical tangent rather than a finite derivative.

$$\begin{array}{c|ccc}x&5&2&3\\\hline f(x)&7&3&8\\f'(x)&4&5&6\end{array}$$

Because f(2)=3, we know f⁻¹(3)=2.

$$(f^{-1})'(3)=\frac1{f'(f^{-1}(3))}=\frac1{f'(2)}=\frac15$$

Taking 1/f′(3)=1/6 would use the wrong column.

$$\theta=\arcsin x,\qquad\sin\theta=x,\qquad-\frac\pi2\le\theta\le\frac\pi2$$

Restrict sine to this interval to obtain a one-to-one function.

$$\cos\theta=\sqrt{1-x^2}$$

On this branch cosine is nonnegative. A unit-circle triangle gives the square root.

$$\frac{d}{dx}\arcsin x=\frac1{\cos(\arcsin x)}=\frac1{\sqrt{1-x^2}}\quad(|x|<1)$$

The endpoint inputs remain in the function's domain, but not in this derivative's domain.

$$\theta=\arccos x,\qquad0\le\theta\le\pi$$

Cosine decreases on this branch, so its inverse also decreases.

$$\frac{d}{dx}\arccos x=\frac1{-\sin(\arccos x)}=-\frac1{\sqrt{1-x^2}}\quad(|x|<1)$$

The unit-circle identity determines the magnitude, and the derivative of cosine supplies the sign.

$$\theta=\arctan x,\qquad\tan\theta=x$$

The inverse derivative rule gives 1/sec²θ.

$$\sec^2\theta=1+\tan^2\theta=1+x^2$$

This identity gives the same result as the unit-circle algebra in the recording.

$$\frac{d}{dx}\arctan x=\frac1{1+x^2}$$

The denominator is positive for every real x.