Derivatives of Inverse Trigs via Implicit Differentiation

We can use implicit differentiation to find derivatives of
inverse
functions. Recall that the equation $$y = f^{-1}(x)$$ means the
same things as
$$x = f(y).$$
Taking derivatives of both sides gives
$$\frac{d}{dx}x=\frac{d}{dx}f(y) \qquad\text{ and using the chain
rule we get }\qquad 1 = f'(y) \frac{dy}{dx}.$$
Dividing both sides by $f'(y)$ (and swapping sides) gives
$$\frac{dy}{dx} = \frac{1}{f'(y)}.$$ Once we rewrite $f'(y)$ in
terms of
$x$, we have the derivative of $f^{-1}(x)$.

In the following video, we use this trick to differentiate
the inverse trig functions $\sin^{-1}$, $\cos^{-1}$ and
$\tan^{-1}$.