Question

Find the derivative of y=log(e^x sin^5 x)

Answer 1

Answer:

$\boxed{\pink{\huge{1 + 5 \: \: cot \: x}}}$

Step-by-step explanation:

$\bold{To \: find}\longrightarrow \\ derivative \: of \: \\ y = log( {e}^{x} {sin}^{5}x )$

$\bold{Concept \: used}\longrightarrow \\ \boxed{\blue{ \dfrac{d}{dx} (x) = 1}} \\ \boxed{\pink{ \dfrac{d}{dx} (logx) = \dfrac{1}{x}}} \\\boxed{\green{ \dfrac{d}{dx} (sinx) = cosx}} \\\boxed{\red{ log {x}^{m} = m \: logx}}$

$solution \\ y = log( {e}^{x} {sin}^{5}x )$

$= > y = log( {e}^{x} ) + log( {sin}^{5} x)$

$= > y = x \: log(e) + 5 \: logsinx$

$= > y = x \: (1) + 5 \: log \: sinx$

$= > y = x + log \: sinx$

$differentiating \: with \: respect \: to \: x$

$= > \dfrac{dy}{dx} = \dfrac{d}{dx} (x) + 5 \dfrac{d}{dx} (logsin5x)$

$= > \dfrac{dy}{dx} = 1 + 5 \: \dfrac{1}{sinx} \dfrac{d}{dx} (sinx)$

$= > \dfrac{dy}{dx} = 1 + 5 \dfrac{1}{sinx} cosx$

$= > \dfrac{dy}{dx} = 1 + 5 \dfrac{cosx}{sinx}$

$= > \dfrac{dy}{dx} = 1 + 5 \: cotx$