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If y = (sin–1 x)2, prove that: (1–x2) y2 – xy1– 2=0

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Note: y2 represents second order derivative i.e.\(\frac{d^2y}{dx^2}\) and y1 = dy/dx

Given,

y = (sin–1 x)2 ……equation 1

to prove : (1–x2) y2–xy1–2=0

We notice a second–order derivative in the expression to be proved so first take the step to find the second order derivative.

Let’s find \(\frac{d^2y}{dx^2}\)

As, \(\frac{d^2y}{dx^2}=\frac{d}{dx}(\frac{dy}{dx})\)

So, lets first find dy/dx

\(\frac{dy}{dx}=\frac{d}{dx}(sin^{-1}x)^2\) 

Using chain rule we will differentiate the above expression

And y = et

Again differentiating with respect to x applying product rule:

Using equation 2 :

∴ (1–x2) y2–xy1–2=0 ……proved

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