If $y=\sin (\sin x)$, prove that $: \frac{d^{2} y}{d x^{2}}+\tan x \cdot \frac{d y}{d x}+y \cos ^{2} x=0$
Given,
$y=\sin (\sin x) \ldots \ldots .$ equation 1
To prove: $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+\tan \mathrm{x} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y} \cos ^{2} \mathrm{x}=0$
We notice a second-order derivative in the expression to be proved so first take the step to find the second order derivative.
Let's find $\frac{d^{2} y}{d x^{2}}$
As $\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{d y}{d x}\right)$
So, lets first find $\mathrm{dy} / \mathrm{dx}$
$\frac{d y}{d x}=\frac{d}{d x} \sin (\sin x)$
Using chain rule, we will differentiate the above expression
Let $t=\sin x \Longrightarrow \frac{d t}{d x}=\cos x$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{dt}} \frac{\mathrm{dt}}{\mathrm{dx}}$
$\frac{d y}{d x}=\cos t \cos x=\cos (\sin x) \cos x \ldots \ldots .$ equation 2
Again differentiating with respect to $x$ applying product rule:
$\frac{d^{2} y}{d x^{2}}=\cos x \frac{d}{d x} \cos (\sin x)+\cos (\sin x) \frac{d}{d x} \cos x$
Using chain rule again in the next step-
$\frac{d^{2} y}{d x^{2}}=-\cos x \cos x \sin (\sin x)-\sin x \cos (\sin x)$
$\frac{d^{2} y}{d x^{2}}=-y \cos ^{2} x-\tan x \cos x \cos (\sin x)$
[using equation $1: y=\sin (\sin x)$ ]
And using equation 2, we have:
$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-y \cos ^{2} \mathrm{x}-\tan \mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}$
$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+\mathrm{y} \cos ^{2} \mathrm{x}+\tan \mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}=0 \ldots \ldots .$ proved