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Question
If y log (1 + cos x), prove that \[\frac{d^3 y}{d x^3} + \frac{d^2 y}{d x^2} \cdot \frac{dy}{dx} = 0\] ?
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Solution
Here,
\[y = \log\left( 1 + \cos x \right)\]
\[\text { Differentiating w . r . t . x, we get }\]
\[\frac{d y}{d x} = \frac{- \sin x}{1 + \cos x}\]
\[\text { Differentiating again w . r . t . x, we get }\]
\[\frac{d^2 y}{d x^2} = \frac{- \cos x - \cos^2 x - \sin^2 x}{\left( 1 + \cos x \right)^2} = \frac{- \left( \cos x + 1 \right)}{\left( 1 + \cos x \right)} = \frac{- 1}{1 + \cos x}\]
\[\text { Differentiating again w . r . t . x, we get }\]
\[\frac{d^3 y}{d x^3} = \frac{- \sin x}{\left( 1 + \cos x \right)^2}\]
\[ \Rightarrow \frac{d^3 y}{d x^3} + \frac{\sin x}{\left( 1 + \cos x \right)^2} = 0\]
\[ \Rightarrow \frac{d^3 y}{d x^3} + \left( \frac{- 1}{1 + \cos x} \right)\left( \frac{- \sin x}{1 + \cos x} \right) = 0\]
\[ \Rightarrow \frac{d^3 y}{d x^3} + \frac{d^2 y}{d x^2} \times \frac{d y}{d x} = 0\]
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