Question

What is the derivative of f(x) = cos^-1(x)?

Answer 1

Determine the derivative of $f\left(x\right)={\cos }^{-1}\left(x\right)$.

Consider $y={\cos }^{-1}\left(x\right)$.

$\Rightarrow \cos \left(y\right)=x\dots \left(1\right)$

Now, differentiate $\left(1\right)$ implicitly with respect to $x$.

$-\sin \left(y\right)\frac{dy}{dx}=1\cdots \left(2\right)$

Apply the trigonometric identity ${\sin }^2\left(y\right)+{\cos }^2\left(y\right)=1$.

Substitute the vallue of $\cos \left(y\right)$ from $\left(1\right)$ into the trigometrric identity.

$\begin{array}{rcl}{\sin }^2\left(y\right)+x^2& =& 1\\ & \Rightarrow & {\sin }^2\left(y\right)=1-x^2\\ & \Rightarrow & \sin \left(y\right)=\sqrt{1-x^2}\end{array}$

Substitute the value of $\sin \left(y\right)$ into $\left(2\right)$.

$\begin{array}{rcl}-\sqrt{1-x^2}\frac{dy}{dx}& =& 1\\ & \Rightarrow & \frac{dy}{dx}=\frac{1}{-\sqrt{1-x^2}}\\ & \Rightarrow & \frac{dy}{dx}=-\frac{1}{\sqrt{1-x^2}}\end{array}$

Hence, the derivative of the given function is $\begin{array}{rcl}\frac{dy}{dx}& =& -\frac{1}{\sqrt{1-x^2}},-1<x<1\end{array}$.