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Explain solution RD Sharma class 12 chapter 10 Differentiation exercise Very short answers question 29 maths

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Answer:

The value of    \frac{d y}{d x} \text { is }-e^{x} \tan e^{x}
Hint:

Differentiating both sides with respect to x.

Given:

y=\log \cos e^{x}
Solution:  

y=\log \cos e^{x}

Differentiating both sides with respect to x.

\frac{d y}{d x}=\frac{d\left\{\log \left(\cos e^{x}\right)\right\}}{d x}

\frac{d y}{d x}=\frac{1}{\cos e^{x}} \cdot \frac{d\left(\cos e^{x}\right)}{d e^{x}} \cdot \frac{d e^{x}}{d x}\left(\therefore \frac{d \log x}{d x}=\frac{1}{x}\right)

=\frac{1}{\cos e^{x}} \cdot\left(-\sin e^{x}\right) \cdot e^{x}\left(\therefore \frac{d e^{x}}{d x}=e^{x}\right)

\begin{aligned} &=\frac{-\sin e^{x}}{\cos e^{x}} \cdot e^{x} \\\\ &=-\tan e^{x} \cdot e^{x} \\\\ &=-e^{x} \cdot \tan e^{x} \end{aligned}

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