#### Question

If *y* = sin (sin *x*), prove that \[\frac{d^2 y}{d x^2} + \tan x \cdot \frac{dy}{dx} + y \cos^2 x = 0\] ?

#### Solution

Here,

\[y = \sin\left( \sin x \right)\]

\[\text { Differentiating w . r . t . x, we get }\]

\[\frac{d y}{d x} = \cos\left( \sin x \right) \cos x\]

\[\text { Differentiating again w . r . t . x, we get }\]

\[\frac{d^2 y}{d x^2} = - \sin\left( \sin x \right) \cos^2 x - \cos\left( \sin x \right) \sin x\]

\[ \Rightarrow \frac{d^2 y}{d x^2} = - \sin\left( \sin x \right) \cos^2 x - \cos\left( \sin x \right) \cos x\tan x\]

\[ \Rightarrow \frac{d^2 y}{d x^2} = - y \cos^2 x - \tan x\frac{d y}{d x}\]

\[ \Rightarrow \frac{d^2 y}{d x^2} + \tan x\frac{d y}{d x} + y \cos^2 x = 0\]

Hence proved.

Is there an error in this question or solution?

Solution If Y = Sin (Sin X), Prove that D 2 Y D X 2 + Tan X ⋅ D Y D X + Y Cos 2 X = 0 ? Concept: Simple Problems on Applications of Derivatives.