# How do you differentiate  y=cos^-1(1-2x^2)?

##### 2 Answers
Mar 6, 2016

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

#### Explanation:

Using the$\textcolor{b l u e}{\text{ chain rule }}$

d/dx[f(g(x)] = f'(g(x)) . g'(x)

and the standard derivative $\frac{d}{\mathrm{dx}} \left({\cos}^{-} 1 x\right) = \frac{- 1}{\sqrt{1 - {x}^{2}}}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 1}{\sqrt{1 - {\left(1 - 2 {x}^{2}\right)}^{2}}} \frac{d}{\mathrm{dx}} \left(1 - 2 {x}^{2}\right)$

$= \frac{4 x}{\sqrt{1 - 1 + 4 {x}^{2} - 4 {x}^{4}}} = \frac{4 x}{\sqrt{4 {x}^{2} \left(1 - {x}^{2}\right)}}$

$= \frac{4 x}{2 x \sqrt{1 - {x}^{2}}} = \frac{2}{\sqrt{1 - {x}^{2}}}$

Mar 6, 2016

$d y = \frac{4 x}{1 - 2 {x}^{2}} ^ 2 {\sin}^{-} 1 \left(1 - 2 {x}^{2}\right) d x$

#### Explanation:

$y = \cos u \text{ } {y}^{'} = - {u}^{'} \cdot \sin u$
$y = {\cos}^{-} 1 \left(1 - 2 {x}^{2}\right) \text{ } y = \cos \left(\frac{1}{1 - 2 {x}^{2}}\right)$
$d y = - \left(\frac{0 \cdot \left(1 - 2 {x}^{2}\right) + 4 x \cdot 1}{1 - 2 {x}^{2}} ^ 2\right) \sin \left(\frac{1}{1 - 2 {x}^{2}}\right) \cdot d x$
$d y = \frac{4 x}{1 - 2 {x}^{2}} ^ 2 {\sin}^{-} 1 \left(1 - 2 {x}^{2}\right) d x$