If y = 10 ( ^ -1 ( 1+x^2 -1 x ))^2 , where x ≠ 0 , then the value...

Given $y = 10 \left(\tan^{-1}\left(\dfrac{\sqrt{1+x^2}-1}{x}\right)\right)^2$

Put $x = \tan\theta$, so $\theta = \tan^{-1}x$.

The inner expression simplifies as:
$\dfrac{\sqrt{1+\tan^2\theta}-1}{\tan\theta} = \dfrac{\sec\theta - 1}{\tan\theta} = \dfrac{1 - \cos\theta}{\sin\theta}$

Using half-angle formulas:
$\dfrac{2\sin^2(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)} = \tan\left(\dfrac{\theta}{2}\right)$

So, $\tan^{-1}\left(\tan\left(\dfrac{\theta}{2}\right)\right) = \dfrac{\theta}{2} = \dfrac{1}{2}\tan^{-1}x$

Substitute this back into the equation for $y$:
$y = 10 \left(\dfrac{1}{2}\tan^{-1}x\right)^2 = \dfrac{10}{4}(\tan^{-1}x)^2 = \dfrac{5}{2}(\tan^{-1}x)^2$

Differentiating with respect to $x$:
$y' = \dfrac{5}{2} \cdot 2\tan^{-1}x \cdot \dfrac{1}{1+x^2} = \dfrac{5\tan^{-1}x}{1+x^2}$

Rearranging gives:
$(1+x^2)y' = 5\tan^{-1}x$

Squaring both sides to avoid the quotient rule:
$(1+x^2)^2 (y')^2 = 25(\tan^{-1}x)^2$

Since $y = \dfrac{5}{2}(\tan^{-1}x)^2$, we have $10y = 25(\tan^{-1}x)^2$. Thus:
$(1+x^2)^2 (y')^2 = 10y$

Differentiating implicitly with respect to $x$ again:
$(1+x^2)^2 \cdot 2y'y'' + (y')^2 \cdot 2(1+x^2)(2x) = 10y'$

Dividing the entire equation by $2y'$ (since $y' \ne 0$ for $x \ne 0$):
$(1+x^2)^2 y'' + 2x(1+x^2)y' = 5$

Answer: $5$