Question

log (x cosx) derivative​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:y \: = \: log(x \: cosx)$

$\rm :\longmapsto\:y \: = \: log(x) + log(cosx)$

$\: \: \: \: \: \: \: \: \: \: \green{\boxed{ \bf{ \: log(xy) = log(x) + log(y) }}}$

On differentiating both sides, w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}y \: = \dfrac{d}{dx}( \: log(x) + log(cosx) )$

$\rm :\longmapsto\:\dfrac{dy}{dx} \: = \dfrac{d}{dx} \: log(x) + \dfrac{d}{dx} log(cosx)$

$\rm :\longmapsto\:\dfrac{dy}{dx} \: = \dfrac{1}{x} \: + \dfrac{1}{cosx} \dfrac{d}{dx}(cosx)$

$\rm :\longmapsto\:\dfrac{dy}{dx} \: = \dfrac{1}{x} \: + \dfrac{1}{cosx} ( - sinx)$

$\bf :\longmapsto\:\dfrac{dy}{dx} \: = \dfrac{1}{x} \: - tanx$

Additional Information :-

$\green{\boxed{ \bf{ \:\dfrac{d}{dx}k = 0}}}$

$\green{\boxed{ \bf{ \:\dfrac{d}{dx}x = 1}}}$

$\green{\boxed{ \bf{ \:\dfrac{d}{dx}sinx = cosx}}}$

$\green{\boxed{ \bf{ \:\dfrac{d}{dx}cosx = - \: sinx}}}$

$\green{\boxed{ \bf{ \:\dfrac{d}{dx}cotx = - \: {cosec}^{2} x}}}$

$\green{\boxed{ \bf{ \:\dfrac{d}{dx}tanx = \: {sec}^{2} x}}}$

$\green{\boxed{ \bf{ \:\dfrac{d}{dx} \sqrt{x} = \dfrac{1}{2 \sqrt{x} } }}}$

$\green{\boxed{ \bf{ \:\dfrac{d}{dx}secx = secx \: tanx}}}$

$\green{\boxed{ \bf{ \:\dfrac{d}{dx}cosecx = - \: cosecx \: cotx}}}$

$\green{\boxed{ \bf{ \:\dfrac{d}{dx}logx = \dfrac{1}{x} }}}$