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

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

Answers (1)

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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