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

Apr 26, 2018

$\frac{d}{\mathrm{dx}} {\cos}^{-} 1 x = - \frac{1}{\sqrt{1 - {x}^{2}}}$

#### Explanation:

In general,

$\frac{d}{\mathrm{dx}} {\cos}^{-} 1 x = - \frac{1}{\sqrt{1 - {x}^{2}}}$

Here's how we obtain this common derivative:

$y = {\cos}^{-} 1 x \to x = \cos y$ from the definition of an inverse function.

Differentiate both sides of $x = \cos y .$

This will entail using Implicit Differentiation on the right side:

$\frac{d}{\mathrm{dx}} \left(x\right) = \frac{d}{\mathrm{dx}} \cos y$

$1 = - \frac{\mathrm{dy}}{\mathrm{dx}} \sin y$

Solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\sin} y$

We need to get rid of the $\sin y .$

We previously said $y = {\cos}^{-} 1 x$. So,

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\sin} \left({\cos}^{-} 1 x\right)$

Now, recall the identity

${\sin}^{2} x + {\cos}^{2} x = 1$

In the identity, replace $x$ with ${\cos}^{-} 1 x :$

${\sin}^{2} \left({\cos}^{-} 1 x\right) + {\cos}^{2} \left({\cos}^{-} 1 x\right) = 1$

${\cos}^{2} \left({\cos}^{-} 1 x\right) = {\left(\cos \left({\cos}^{-} 1 x\right)\right)}^{2} = {x}^{2}$

${\sin}^{2} \left({\cos}^{-} 1 x\right) + {x}^{2} = 1$

${\sin}^{2} \left({\cos}^{-} 1 x\right) = 1 - {x}^{2}$

$\sin \left({\cos}^{-} 1 x\right) = \sqrt{1 - {x}^{2}}$

Thus,

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\sqrt{1 - {x}^{2}}}$

Apr 26, 2018

$f \left(x\right) = {\cos}^{-} 1 \left(x\right) \text{ "=>" } \cos \left(f \left(x\right)\right) = x$

Take the derivative of both sides. Use the chain rule on the left.

$- \sin \left(f \left(x\right)\right) \cdot f ' \left(x\right) = 1$

$\implies \text{ } f ' \left(x\right) = \frac{- 1}{\sin} \left(f \left(x\right)\right) = \frac{- 1}{\sqrt{1 - {\cos}^{2} \left(f \left(x\right)\right)}}$

The last step came from the identity ${\sin}^{2} \left(\theta\right) + {\cos}^{2} \left(\theta\right) = 1$, which is restated as $\sin \left(\theta\right) = \sqrt{1 - {\cos}^{2} \left(\theta\right)}$. We should also remember that $\cos \left(f \left(x\right)\right) = x$, which we saw in the first line, so finally:

$f ' \left(x\right) = \frac{- 1}{\sqrt{1 - {x}^{2}}}$

Note about domain: the domain of ${\cos}^{-} 1 \left(x\right)$ is $0 < = x < = \pi$. Note that on this interval, $\sin \left(x\right) > = 0$. This allows us to only take the positive root when we say that $\sin \left(f \left(x\right)\right) = \sqrt{1 - {\cos}^{2} \left(f \left(x\right)\right)}$.