If $y = \sin^{-1}x$, show that $(1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} = 0$.

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Accepted Answer

>

> [!TLDR]
> Differentiate $y=\sin^{-1}x$ once to get $y'$, write it as $(1-x^2)^{1/2}y'=1$, then differentiate again.
>
> $y=\sin^{-1}x \Rightarrow y'=\dfrac{1}{\sqrt{1-x^2}}$, i.e. $\sqrt{1-x^2}\,y'=1$.
>
> Differentiate both sides w.r.t. $x$:
>
> $\sqrt{1-x^2}\,y'' + y'\cdot\dfrac{-x}{\sqrt{1-x^2}} = 0$
> …

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.If $y = \sin^{-1}x$ , show that $(1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} = 0$ .

Concept understanding — Successive Differentiation

Successive Differentiation — Repeated Slopes

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Q.If y=sin⁡−1xy = \sin^{-1}xy=sin−1x, show that (1−x2)d2ydx2−xdydx=0(1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} = 0(1−x2)dx2d2y​−xdxdy​=0.

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Q.If $y = \sin^{-1}x$ , show that $(1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} = 0$ .

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