Question

Differentiate:log(3+x)​

Answer 1

$\frac{1}{3 + x}$

Explanation:

$Let, \: \: \: y = log(3 + x) \\ Differentiating \: \: with \: \: respect \: \: to \: \: x, \\ \frac{dy}{dx} = \frac{d \{ log(3 + x\}}{dx} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{d \{ log(3 + x\}}{d(3 + x)} \times \frac{d(3 + x)}{dx} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1}{(3 + x)} \times (0 + 1) \\ \: \: = \frac{1}{(3 + x)} \times 1 \\ = \boxed{ \frac{1}{(3 + x)}}$