Problem Statement
If $y = (\tan^{-1} x)^2$, show that:
$$(x^2 + 1)^2 y_2 + 2x(x^2 + 1) y_1 = 2$$
where $y_1 = \frac{dy}{dx}$ and $y_2 = \frac{d^2 y}{dx^2}$.
Verified Solution & Marking Scheme
Find first derivative y₁
$$y = (\tan^{-1} x)^2 \implies y_1 = 2(\tan^{-1} x) \cdot \frac{1}{1 + x^2}$$
Multiply both sides by $$(1 + x^2)$$:
$$(1 + x^2) y_1 = 2\tan^{-1} x$$
Differentiate both sides with respect to x using product rule
$$\frac{d}{dx}[(1 + x^2) y_1] = \frac{d}{dx}[2\tan^{-1} x]$$
$$(1 + x^2) y_2 + 2x y_1 = 2 \cdot \frac{1}{1 + x^2}$$
Multiply across by (1 + x²)
$$(1 + x^2)^2 y_2 + 2x(1 + x^2) y_1 = 2$$ (Hence proved).