5. Derivatives of Inverse (Trigonometric) Functions

Let’s take the equation of $y=\arcsin (x)$. From previous sections, we do know that $y=\arcsin (x)$ is equal to $\sin (y)=x$.

Info

$y=\arcsin (x)$ and $\sin (y)=x$

$y=\arccos (x)$ and $\cos (y)=x$

$y=\arctan (x)$ and $\tan (y)=x$

Remember that $y$ is equal to the function. Then, the derivative of that would become $\cos (y)\frac{dy}{dx}=1$ which simplifies to $\frac{dy}{dx}=\frac{1}{\cos (y)}=\frac{1}{\cos (\arcsin (x))}=\frac{1}{\sqrt{1-x^2}}$. This pertains the information below.

Info

$\frac{d}{dx}\arcsin (x)=\frac{1}{\sqrt{1-x^2}}$

$\frac{d}{dx}\arccos (x)=\frac{-1}{\sqrt{1-x^2}}$

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

This will be the same information for any other similarly to $\arccos$ and $arctan$.