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Differentiate Log √ 1 − Cos X 1 + Cos X ? - Mathematics

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प्रश्न

Differentiate \[\log \sqrt{\frac{1 - \cos x}{1 + \cos x}}\] ?

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उत्तर

\[\text{Let } y = \log\sqrt{\frac{1 - \cos x}{1 + \cos x}}\]

\[ \Rightarrow y = \log \left( \frac{1 - \cos x}{1 + \cos x} \right)^\frac{1}{2} \]

\[ \Rightarrow y = \frac{1}{2}\log\left( \frac{1 - \cos x}{1 + \cos x} \right) \left[ \text{using} \log a^b = b\log a \right]\]

\[\text{Differentiate it with respect to x we get}, \]

\[\frac{d y}{d x} = \frac{d}{dx}\left\{ \frac{1}{2}\log\left( \frac{1 - \cos x}{1 + \cos x} \right) \right\}\] 

\[ = \frac{1}{2} \times \frac{1}{\left( \frac{1 - \cos x}{1 + \cos x} \right)} \times \frac{d}{dx}\left( \frac{1 - \cos x}{1 + \cos x} \right) \left[ \text{Using chain rule} \right]\]

\[ = \frac{1}{2}\left( \frac{1 + \cos x}{1 - \cos x} \right)\left[ \frac{\left( 1 + \cos x \right)\frac{d}{dx}\left( 1 - \cos x \right) - \left( 1 - \cos x \right)\frac{d}{dx}\left( 1 + \cos x \right)}{\left( 1 + cos x \right)^2} \right] \left[ \text{Using quotient rule} \right]\]

\[ = \frac{1}{2}\left( \frac{1 + \cos x}{1 - \cos x} \right)\left[ \frac{\left( 1 + \cos x \right)\left( \sin x \right) - \left( 1 - \cos x \right)\left( - \sin x \right)}{\left( 1 + \cos x \right)^2} \right]\]

\[ = \frac{1}{2}\left( \frac{1 + \cos x}{1 - \cos x} \right)\left[ \frac{\sin x + \sin x \cos x + \sin x - \sin x \cos x}{\left( 1 + \cos x \right)^2} \right]\]

\[ = \frac{1}{2}\left( \frac{1 + \cos x}{1 - \cos x} \right)\left[ \frac{2\sin x}{\left( 1 + \cos x \right)^2} \right]\]

\[ = \frac{\sin x}{\left( 1 - \cos x \right)\left( 1 + \cos x \right)}\]

\[ = \frac{\sin x}{1 - \cos^2 x}\]

\[ = \frac{\sin x}{\sin^2 x} \]

\[ = \frac{1}{\sin x}\]

`= "cosec "x`

\[So, \frac{d}{dx}\left( \log\sqrt{\frac{1 - \cos x}{1 + \cos x}} \right) = cosec x\]

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अध्याय 11: Differentiation - Exercise 11.02 [पृष्ठ ३७]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 11 Differentiation
Exercise 11.02 | Q 26 | पृष्ठ ३७

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