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If Y = Sin (Sin X), Prove that D 2 Y D X 2 + Tan X ⋅ D Y D X + Y Cos 2 X = 0 ? - Mathematics

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प्रश्न

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

उत्तर

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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पाठ 12: Higher Order Derivatives - Exercise 12.1 [पृष्ठ १७]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 12 Higher Order Derivatives
Exercise 12.1 | Q 18 | पृष्ठ १७

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