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  • explain solution RD Sharma class 12 chapter Differentiation exercise 10.2 question 39 maths

explain solution RD Sharma class 12 chapter Differentiation exercise 10.2 question 39 maths

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Answer:  \frac{2^{x}}{\left(x^{2}+3\right)^{2}}\left[\cos x \log _{e} 2-\sin x-\frac{4 x \cos x}{\left(x^{2}+3\right)}\right]

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

Given:   \frac{2^{x} \cos x}{\left(x^{2}+3\right)^{2}}

Solution:

Let  y=\frac{2^{x} \cos x}{\left(x^{2}+3\right)^{2}}

Differentiate with respect to x,

\frac{d y}{d x}=\frac{d}{d x}\left[\frac{2^{x} \cos x}{\left(x^{2}+3\right)^{2}}\right]

\frac{d y}{d x}=\left[\frac{\left(x^{2}+3\right)^{2} \frac{d}{d x}\left(2^{x} \cos x\right)-\left(2^{x} \cos x\right) \frac{d}{d x}\left(x^{2}+3\right)^{2}}{\left[\left(x^{2}+3\right)^{2}\right]^{2}}\right]... \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}

\frac{d y}{d x}=\left[\frac{\left(x^{2}+3\right)^{2}\left\{2^{x} \frac{d}{d x} \cos x+\cos x \frac{d}{d x} 2^{x}\right\}-\left(2^{x} \cos x\right) 2\left(x^{2}+3\right) \frac{d}{d x}\left(x^{2}+3\right)}{\left[x^{2}+3\right]^{4}}\right]

\frac{d y}{d x}=\left[\frac{\left(x^{2}+3\right)^{2}\left\{-2^{x} \sin x+\cos x 2^{x} \log _{e} 2\right\}-2\left(2^{x} \cos x\right)\left(x^{2}+3\right)(2 x)}{\left[x^{2}+3\right]^{4}}\right]

\frac{d y}{d x}=\left[\frac{2^{x}\left(x^{2}+3\right)^{2}\left\{\left(\cos x \log _{e} 2-\sin x\right\}-\frac{4 x \cos x}{\left(x^{2}+3\right)}\right.}{\left[x^{2}+3\right]^{4}}\right]

\frac{d y}{d x}=\frac{2^{x}}{\left(x^{2}+3\right)^{2}}\left[\cos x \log _{e} 2-\sin x-\frac{4 x \cos x}{\left(x^{2}+3\right)}\right]

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