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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.
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