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

Question

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

Accepted Answer

> [!TLDR]
> Differentiating $y=(	an^{-1}x)^2$ twice and clearing denominators yields the required relation.

1. Given $y=(	an^{-1}x)^2$. Differentiate once, using the chain rule with $\dfrac{d}{dx}(	an^{-1}x)=\dfrac{1}{1+x^2}$:
$$y_1=2	an^{-1}x\cdot\frac{1}{1+x^2}=\frac{2	an^{-1}x}{1+x^2}.$$

2. Clear the denominator:
$$(1+x^2)\,y_1=2	an^{-1}x.\qquad(\star)$$

3. Differentiate $(\star)$ with respect to $x$. Left side by the product rule, right side using $\dfrac{d}{dx}(	an^{-1}x)=\dfrac{1}{1+x^2}$:
$$(1+x^2)\,y_2+y_1\cdot(2x)=2\cdot\frac{1}{1+x^2}.$$

4. That is, …

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

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

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