CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

$\text{x}=\cos\theta,\text{y}=\sin^3$

Differentiating w.r.t.x, we get

$\frac{\text{dx}}{\text{d}\theta}=-\sin\theta\ \text{and}\ \frac{\text{dy}}{\text{d}\theta}=3\sin^2\theta\cos\theta$

$\therefore\frac{\text{dy}}{\text{dx}}=\frac{3\sin^2\theta\cos\theta}{-\sin\theta}=-3\sin\theta\cos\theta$

Differentiating w.r.t.x, we get

$\frac{\text{d}^2\text{y}}{\text{dx}^2}=(-3\cos^2\theta+3\sin^2\theta)\frac{\text{d}\theta}{\text{dx}}\frac{)-3\cos^2\theta+3\sin^2\theta)}{-\sin\theta}$

Now,

$\text{LHS}=\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2$

$=\sin^3\theta\times\frac{(-3\cos^2\theta+3\sin^2\theta)}{\sin\theta}+(-3\sin\theta\cos\theta)^2$

$=3\sin^2\theta\cos^2\theta-3\sin^4\theta+9\sin^2\theta\cos^2\theta$

$=12\sin^2\theta\cos^2\theta-3\sin^4\theta$

$=3\sin^2\theta(4\cos^2\theta-\sin^2\theta)$

$=3\sin^2\theta(5\cos^2\theta-1)$

$=\text{RHS}$