Question

If y=sin(sinx),prove y₂+tanx.y₁+ycos²x=0

Answer 1

Answer:

derive two times and substitute the value you get in the equation

Answer 2

Given:

$y=\sin (\sin x)$

Step-by-step explanation:

On differentiating the given equation wrt to ‘x’, we get,

$y_{1}=\frac{d y}{d x}=\cos (\sin x) \cos x$

$\bold{[\because \frac{d y}{d x}(x y)=x y^{\prime}+x^{\prime} y]}$

Again on differentiating the above equation wrt to ‘x’, we get,

$y_{2}=\frac{d^{2} y}{d x^{2}}=-\sin x \cos (\sin x)+\cos x(-\sin (\sin x) \cos x)$

$\Rightarrow y_{2}=-\sin x \cos (\sin x)-\cos ^{2} x \sin (\sin x)$

$\Rightarrow y_{2}=-\sin x \cos (\sin x)-y \cos ^{2} x$

On multiplying and diving $\cos{x}$, we get,

$\Rightarrow y_{2}=-\frac{\sin x}{\cos x} \cos (\sin x) \cos x-y \cos ^{2} x$

$\Rightarrow y_{2}=-\frac{\sin x}{\cos x} y_{1}-y \cos ^{2} x$

$\Rightarrow y_{2}=-y_{1} \tan x-y \cos ^{2} x$

$\therefore y_{2}+\tan x y_{1}+y \cos ^{2} x=0$

Hence proved.