If $y=\sin^{-1}x$, then prove that $(1-x^2)\dfrac{d^2y}{dx^2}=x\dfrac{dy}{dx}$.

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Accepted Answer

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> [!TLDR]
> Square the first derivative to clear the root, differentiate again, and cancel $y'$ to obtain $(1-x^2)y''=xy'$.
>
> **Concept.** To avoid differentiating a square root twice, square the relation $y'=\frac{1}{\sqrt{1-x^2}}$ first.
>
> **Step 1.** $y=\sin^{-1}x\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-x^2}}$, so $(1-x^2)\left(\dfrac{dy}{dx}\right)^2=1$.
>
> **Step 2 — differentiate w.r.t. $x$:**
> $$-2x\left(\frac{dy}{dx}\right)^2+(1-x^2)\cdot 2\,\frac{dy}{dx}\,\frac{d^2y}{dx^2}=0.$$
> …

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

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

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Q.If $y=\sin^{-1}x$ , then prove that $(1-x^2)\dfrac{d^2y}{dx^2}=x\dfrac{dy}{dx}$ .

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