Question

differtentian of ln(1+x)^1+x

Answer 1

$y = { ln(1 + x) }^{1 + x} \\ taking \: log \: both \: sides \\ ln(y) = ln( {ln(1 + x)}^{1 + x} ) \\ ln(y) = (1 + x)( ln( ln(1 + x) ) \\ diffrentiating \: both \: sides \: wrt \: x \\ \frac{1}{y} \frac{dy}{dx} = ln( ln(1 + x) ) + (1 + x) \times \frac{1}{ ln(1 + x) } \times \frac{1}{1 + x} \\ \frac{dy}{dx} = y( ln( ln(1 + x) ) + \frac{1}{ ln(1 + x) }$
hope it will help you.