Question

If $y=\cos \left(\sin x\right)$, show that:
$\frac{d^{2}y}{d{x}^2}+\tan x\frac{dy}{dx}+y{\cos }^{2}x=0$

Answer 1

Given $y=\cos \left(\sin x\right)$

${\cos }^{-1}y=\sin x$

Differentiating wrt $x$ both sides we get ,

$\frac{-1}{\sqrt{1-y^2}}\frac{dy}{dx}=\cos x$

$\frac{dy}{dx}=-\sqrt{1-y^2}\cos x$

Differentiating wrt $x$ both sides we get ,

$\frac{d^{2}y}{d{x}^2}=\sqrt{1-y^2}\sin x+\cos x\frac{y}{\sqrt{1-y^2}}\frac{dy}{dx}$

$\frac{d^{2}y}{d{x}^2}=\sin x\sqrt{1-y^2}-\cos x\frac{y}{\sqrt{1-y^2}}\sqrt{1-y^2}\cos x$

$\frac{d^{2}y}{d{x}^2}=\sqrt{1-y^2}\sin x-y{\cos }^{2}x$

L.H.S $=$

$\frac{d^{2}y}{d{x}^2}+\tan x\frac{dy}{dx}+y{\cos }^{2}x=\sqrt{1-y^2}\sin x-y{\cos }^{2}x+\frac{\sin x}{\cos x}(-\sqrt{1-y^2}\cos x)+y{\cos }^{2}x$

$\frac{d^{2}y}{d{x}^2}+\tan x\frac{dy}{dx}+y{\cos }^{2}x=0$

$=$ R.H.S