Prove that $(1 - x^2)\dfrac{d^2y}{dx^2} = x\dfrac{dy}{dx}$ if $y = \sin^{-1}x$.

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Accepted Answer

>

> [!TLDR]
> From √(1−x²)·y' = 1, differentiating gives (1−x²)y'' = x y'.
>
> Given y = sin⁻¹x.
>
> Step 1: y' = dy/dx = 1/√(1 − x²). So √(1 − x²) · y' = 1.  ...(A)
>
> Step 2: Differentiate (A) with respect to x (product rule):
> √(1−x²)·y'' + y'· d/dx[√(1−x²)] = 0.
> Since d/dx √(1−x²) = −x/√(1−x²):
> √(1−x²)·y'' + y'·(−x/√(1−x²)) = 0. …

Order	Leibniz	Lagrange	Newton
1st	$\frac{dy}{dx}$	$f'(x)$	$\dot{y}$
2nd	$\frac{d^2y}{dx^2}$	$f''(x)$	$\ddot{y}$
$n$ th	$\frac{d^ny}{dx^n}$	$f^{(n)}(x)$	—

Q.Prove that $(1 - x^2)\dfrac{d^2y}{dx^2} = x\dfrac{dy}{dx}$ if $y = \sin^{-1}x$ .

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Q.Prove that (1−x2)d2ydx2=xdydx(1 - x^2)\dfrac{d^2y}{dx^2} = x\dfrac{dy}{dx}(1−x2)dx2d2y​=xdxdy​ if y=sin⁡−1xy = \sin^{-1}xy=sin−1x.

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Q.Prove that $(1 - x^2)\dfrac{d^2y}{dx^2} = x\dfrac{dy}{dx}$ if $y = \sin^{-1}x$ .

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