# How do you find the derivative of the function: (arccosx)/(x+1)?

##### 1 Answer
Dec 19, 2016

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

#### Explanation:

Step 1: Determine the derivative of $y = \arccos x$

The notation $y = \arccos x$ signifies that $\cos y = x$. Therefore:

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

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

We know that ${\sin}^{2} y + {\cos}^{2} y = 1 \to {\sin}^{2} y = 1 - {\cos}^{2} y$ and that $\sin y = \sqrt{1 - {\cos}^{2} y}$. We saw above that if $y = \arccos x$, then $\cos y = x$, so $\sin y = \sqrt{1 - {x}^{2}}$.

The derivative is hence $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\sqrt{1 - {x}^{2}}}$.

Step 2: Determine the derivative of the entire function

Use the quotient rule.

Call the function $f \left(x\right)$:

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

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

Hopefully this helps!