If $y=( an^{-1}x)^2$, then show that $(x^2+1)^2y_2+2x(x^2+1)y_1-2=0$.

Question

If $y=(	an^{-1}x)^2$, then show that $(x^2+1)^2y_2+2x(x^2+1)y_1-2=0$.

Accepted Answer

>

> [!TLDR]
> Two differentiations of $y=(	an^{-1}x)^2$, followed by multiplying through by $(1+x^2)$, produce the required identity.
>
> Let $y=(	an^{-1}x)^2$.
>
> **First derivative** (chain rule):
> $$y_1=\frac{dy}{dx}=2(	an^{-1}x)\cdot\frac{1}{1+x^2}=\frac{2	an^{-1}x}{1+x^2}$$
>
> Rearranging:
> $$(1+x^2)\,y_1=2	an^{-1}x \qquad 	ext{...(i)}$$
>
> **Differentiate (i) again with respect to $x$** (product rule on the LHS):
> $$\frac{d}{dx}\big[(1+x^2)y_1\big]=\frac{d}{dx}\big[2	an^{-1}x\big]$$
> $$(1+x^2)y_2+y_1\cdot2x=\frac{2}{1+x^2}$$
> …

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 $y=(\tan^{-1}x)^2$ , then show that $(x^2+1)^2y_2+2x(x^2+1)y_1-2=0$ .

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Q.If y=(tan⁡−1x)2y=(\tan^{-1}x)^2y=(tan−1x)2, then show that (x2+1)2y2+2x(x2+1)y1−2=0(x^2+1)^2y_2+2x(x^2+1)y_1-2=0(x2+1)2y2​+2x(x2+1)y1​−2=0.

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Q.If $y=(\tan^{-1}x)^2$ , then show that $(x^2+1)^2y_2+2x(x^2+1)y_1-2=0$ .

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