Question

If y = ( sin^-1x )^2 then find d^2y/dx^2

Answer 1

Questìon :

if y = (sin⁻¹x)² , then find d²y/dx²

Theory :

${\red{\boxed{\large{\bold{Chain\:Rule}}}}}$

Let y=f(t) ,t = g(u) and u =m(x) ,then

$\sf\:\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{du} \times \dfrac{du}{dx}$

Solution:

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

Differentiate it with respect to x

$\sf \: \dfrac{dy}{dx} = \dfrac{d(\sin {}^{ - 1} x) {}^{2} }{x}$

$\implies \sf \: \dfrac{dy}{dx} = 2(\sin {}^{ - 1} x )\times \dfrac{1}{ \sqrt{1 - x {}^{2} } }$

$\implies \sf \: \dfrac{dy}{dx} = \dfrac{ 2 \sin {}^{ - 1}x }{ \sqrt{1 - x {}^{2} } }$

Now Differentiate again it with respect to x by ,using Quotient Rule.

$\implies \sf \: \dfrac{d {}^{2} y}{dx {}^{2} } = \dfrac{ \sqrt{1 - x {}^{2} } \times 2 \times \frac{1}{ \sqrt{1 - {x}^{2} } } -2 (\sin {}^{2}x ) \times \frac{1}{2 \sqrt{1 - x {}^{2} } } - 2x}{( \sqrt{1 - x {}^{2} } ) {}^{2} }$

$\implies \bf \: \dfrac{d {}^{2} y}{dx {}^{2} } = \dfrac{2 + \frac{2x( \sin {}^{ - 1}x) }{ \sqrt{1 - x {}^{2} } } }{1 - x {}^{2} }$

It is the required solution!

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More About Differention :

•Quotient rule

Let u = f(x) and g(x).Then ,

$\implies \sf \: \dfrac{d u}{dv } = \dfrac{v \times \frac{du}{dx} - u \times \frac{dv}{dx} }{v {}^{2} }$