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need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.2 question 23

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Answer: 3 e^{\tan 3 x} \sec ^{2} 3 x

Hint: You must know the rules of solving derivative of exponential and trigonometric

Given: e^{\tan 3 x}


Let  y=e^{\tan 3 x}

Differentiating with respect to x

\frac{d y}{d x}=\frac{d}{d x}\left(e^{\tan 3 x}\right)

\frac{d y}{d x}=e^{\tan 3 x} \frac{d}{d x}(\tan 3 x)                         \left[\therefore \frac{d}{d x} e^{x}=e^{x}\right] e^{a x}=e^{x} \frac{d}{d x}(a)

\frac{d y}{d x}=e^{\tan 3 x} \sec ^{2} 3 x \times \frac{d}{d x}(3 x)                \left[\therefore \frac{d}{d x} \tan a x=s \sec ^{2} a x\right]

\begin{aligned} &\frac{d y}{d x}=e^{\tan 3 x} \cdot \sec ^{2} 3 x \times 3 \\ &\frac{d y}{d x}=3 e^{\tan 3 x} \sec ^{2} 3 x \end{aligned}                    \left[\therefore \frac{d}{d x} \tan a x=a \sec ^{2} a x\right]

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