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# Differentiate

Question:

Differentiate

$\left(\frac{e^{x}+\sin x}{1+\log x}\right)$

Solution:

To find: Differentiation of $\left(\frac{e^{x}+\sin x}{1+\log x}\right)$

Formula used: $(i)\left(\frac{u}{v}\right)^{\prime}=\frac{u^{\prime} v-u v^{\prime}}{v^{2}}$ where $v \neq 0$ (Quotient rule)

(ii) $\frac{d \sin x}{d x}=\cos x$

(iii) $\frac{d \log x}{d x}=\frac{1}{x}$

(iv) $\frac{d e^{x}}{d x}=e^{x}$

Let us take $u=\left(e^{x}+\sin x\right)$ and $v=(1+\log x)$

$u^{\prime}=\frac{d u}{d x}=\frac{d\left(e^{x}+\sin x\right)}{d x}=\left(e^{x}+\cos x\right)$

$v^{\prime}=\frac{d v}{d x}=\frac{d(1+\log x)}{d x}=\frac{1}{x}$

Putting the above obtained values in the formula:-

$\left(\frac{u}{v}\right)^{\prime}=\frac{u^{\prime} v-u v^{\prime}}{v^{2}}$ where $v \neq 0$ (Quotient rule)

$\left[\frac{e^{x}+\sin x}{1+\log x}\right]=\frac{\left(e^{x}+\cos x\right) \times(1+\log x)-\left(e^{x}+\sin x\right) \times\left(\frac{1}{x}\right)}{(1+\log x)^{2}}$

$=\frac{x\left(e^{x}+\cos x\right)(1+\log x)-\left(e^{x}+\sin x\right)}{x(1+\log x)^{2}}$

Ans $)=\frac{x\left(e^{x}+\cos x\right)(1+\log x)-\left(e^{x}+\sin x\right)}{x(1+\log x)^{2}}$