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 $\tan^{-1}$.
| \begin{eqnarray*} \frac{d}{dx} \sin^{-1}(x)&=& \frac{1}{\sqrt{1-x^2}}, \cr &&\\ \frac{d}{dx}\cos^{-1}(x)&=&\frac{-1}{\sqrt{1-x^2}}, \cr &&\\ \frac{d}{dx}\tan^{-1}(x)&=&\frac{1}{1+x^2}. \end{eqnarray*} |
You should learn these derivatives; however, they will be most
helpful in the second semester of calculus.