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need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.2 question 31

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Answer: \frac{-8}{\left(e^{2 x}-e^{-2 x}\right)^{2}}

Hint: You must know the rules of solving derivative of exponential function.

Given:  \frac{e^{2 x}+e^{-2 x}}{e^{2 x}-e^{-2 x}}


Let  y=\frac{e^{2 x}+e^{-2 x}}{e^{2 x}-e^{-2 x}}


Differentiate with respect to x,

\frac{d y}{d x}=\frac{d}{d x}\left[\frac{e^{2 x}+e^{-2 x}}{e^{2 x}-e^{-2 x}}\right]

\frac{d y}{d x}=\left\{\frac{\left(e^{2 x}-e^{-2 x}\right) \frac{d}{d x}\left(e^{2 x}+e^{-2 x}\right)-\left(e^{2 x}+e^{-2 x}\right) \frac{d}{d x}\left(e^{2 x}-e^{-2 x}\right)}{\left(e^{2 x}+e^{-2 x}\right)^{2}}\right\}\cdot \cdot \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}


Using quotient rule,

\frac{d y}{d x}=\frac{\left(e^{2 x}-e^{-2 x}\right)\left(2 e^{2 x}-2 e^{-2 x}\right)-\left(e^{2 x}+e^{-2 x}\right)\left(2 e^{2 x}+2 e^{-2 x}\right)}{\left(e^{2 x}+e^{-2 x}\right)^{2}}

\frac{d y}{d x}=\frac{2\left(e^{4 x}+e^{-4 x}-2 e^{2 x} e^{-2 x}-e^{4 x}-e^{-4 x}-2 e^{2 x} e^{-2 x}\right)}{\left(e^{2 x}+e^{-2 x}\right)^{2}}

\frac{d y}{d x}=\frac{-8}{\left(e^{2 x}+e^{-2 x}\right)^{2}}

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