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

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

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Answer 1 · 2018-04-27T10:43:18

$y = {cos}^{- 1} (a x) \Rightarrow \frac{d y}{d x} = - \frac{1}{\sqrt{1 - {(a x)}^{2}}} \frac{d}{d x} (a x)$ $y=cos^-1(ax)=>(dy)/(dx)=-1/sqrt(1-(ax)^2)d/(dx)(ax)$
$\frac{d y}{d x} = - \frac{1}{\sqrt{1 - a^{2} x^{2}}} \times a = \frac{- a}{\sqrt{1 - a^{2} x^{2}}}$ $(dy)/(dx)=-1/sqrt(1-a^2x^2)xxa=(-a)/sqrt(1-a^2x^2)$
Note: $\frac{d}{d X} ({cos}^{- 1} X) = - \frac{1}{\sqrt{1 - X^{2}}}$ $color(red)(d/(dX)(cos^-1X)=-1/sqrt(1-X^2)$

Explanation:

Here,

$y = {cos}^{- 1} (a x)$ $y=cos^-1(ax)$

Let, $u = a x \Rightarrow \frac{d u}{d x} = a$ $u=ax=>(du)/(dx)=a$

$S o, y = {cos}^{- 1} u \Rightarrow \frac{d y}{d u} = - \frac{1}{\sqrt{1 - u^{2}}}$ $So, y=cos^-1u=>(dy)/(du)=-1/sqrt(1-u^2)$

$Using Chain Rule$ $"Using "color(blue)"Chain Rule"$ ,

$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$ $color(blue)((dy)/(dx)=(dy)/(du)*(du)/(dx)$

$:.(dy)/(dx)=-1/sqrt(1-u^2)xxa=(-a)/sqrt(1-u^2),where, u=ax$

$=>(dy)/(dx)=(-a)/sqrt(1-(ax)^2)$

$=>(dy)/(dx)=(-a)/sqrt(1-a^2x^2)$

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

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