Question

x log x
y =
x + log x​

Answer 1

HEYY MATE

$y = \frac{x \: logx}{x + logx}$

$\frac{dy}{dx} = \frac{x \times \frac{1}{x} + \: log \: x \: }{x + \: log \: x} - \frac{x \: log \: x}{ {(x + logx)}^{2} } (1 + \frac{1}{x)}$

$= \frac{(x + logx)(1 + logx) - logx(1 + x)}{ {x + logx}^{2} }$

$= \: \frac{x + ( {logx)}^{2} }{ {(x + logx)}^{2} }$

$\huge\star\sf\red{hope \: it \: helps}\star$