# Find the differentiation of y=cos^-1(ax)?

Apr 27, 2018

$y = {\cos}^{-} 1 \left(a x\right) \implies \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\sqrt{1 - {\left(a x\right)}^{2}}} \frac{d}{\mathrm{dx}} \left(a x\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\sqrt{1 - {a}^{2} {x}^{2}}} \times a = \frac{- a}{\sqrt{1 - {a}^{2} {x}^{2}}}$
Note: color(red)(d/(dX)(cos^-1X)=-1/sqrt(1-X^2)

#### Explanation:

Here,

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

Let, $u = a x \implies \frac{\mathrm{du}}{\mathrm{dx}} = a$

$S o , y = {\cos}^{-} 1 u \implies \frac{\mathrm{dy}}{\mathrm{du}} = - \frac{1}{\sqrt{1 - {u}^{2}}}$

$\text{Using "color(blue)"Chain Rule}$,

color(blue)((dy)/(dx)=(dy)/(du)*(du)/(dx)

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\sqrt{1 - {u}^{2}}} \times a = \frac{- a}{\sqrt{1 - {u}^{2}}} , w h e r e , u = a x$

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

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