Question

find dy /dx if y= sin^-1 (x^2)
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Answer 1

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$\longmapsto \sf y = { \sin}^{ - 1}( {x}^{2})$

$\longmapsto \sf\sin y = {x}^{2}$

$\longmapsto \sf \cos y \bigg\lgroup \frac{dy}{dx} \bigg \rgroup = 2x$

$\longmapsto \sf\frac{dy}{dx} = \frac{2x}{ \cos y}$

$\longmapsto \sf\frac{dy}{dx} = \frac{2x}{ \sqrt{1 - { \sin}^{2}y } }$

$\longmapsto \sf\frac{dy}{dx} = \frac{2x}{ \sqrt{1 - { {(x}^{2}) }^{2} } }$

$\longmapsto\boxed{ \sf\frac{dy}{dx} = \frac{2x}{ \sqrt{1 - {x}^{4} }} }$

Answer 2

Step-by-step explanation:

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$\longmapsto \sf y = { \sin}^{ - 1}( {x}^{2})$

$\longmapsto \sf\sin y = {x}^{2}$

$\longmapsto \sf \cos y \bigg\lgroup \frac{dy}{dx} \bigg \rgroup = 2x$

$\longmapsto \sf\frac{dy}{dx} = \frac{2x}{ \cos y}$

$\longmapsto \sf\frac{dy}{dx} = \frac{2x}{ \sqrt{1 - { \sin}^{2}y } }$

$\longmapsto \sf\frac{dy}{dx} = \frac{2x}{ \sqrt{1 - { {(x}^{2}) }^{2} } }$

$\longmapsto\boxed{ \sf\frac{dy}{dx} = \frac{2x}{ \sqrt{1 - {x}^{4} }} }$