Please solve RD Sharma class 12 chapter Differentiation exercise 10.2 question 29 maths textbook solution

Answer: $e^{x} x^{-2}\left[\frac{1}{x}+\log x-\frac{2}{x} \log x\right]$

Hint: You must know about the rules of solving derivative of exponential and logarithm functions.

Given: $\frac{e^{x} \log x}{x^{2}}$

Solution:

Let  $y=\frac{e^{x} \log x}{x^{2}}$

Differentiate with respect to x,

$\frac{d y}{d x}=\frac{x^{2} \frac{d}{d x}\left(e^{x} \log x\right)-\left(e^{x} \log x\right) \frac{d}{d x} x^{2}}{\left(x^{2}\right)^{2}} \cdot \cdot \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$            [using quotient rule]

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

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

$\frac{d y}{d x}=\frac{x^{2} e^{x}\left\{\frac{(1+x \log x)}{x}-2 x e^{x} \log x\right\}}{x^{4}}$

$\frac{d y}{d x}=\frac{x e^{x}\{1+x \log x-2 \log x\}}{x^{4}}$

$\frac{d y}{d x}=\frac{x e^{x}}{x^{3}}\left[\frac{1}{x}+\frac{x \log x}{x}-\frac{2 \log x}{x}\right]$

$\frac{d y}{d x}=e^{x} x^{-2}\left[\frac{1}{x}+\log x-\frac{2}{x} \log x\right]$