Question

differentiate the function y=9/sinx+3cotx+a^x

answer me fast please​

Answer 1

$\bold{Refer\; to\; the\; attachment\; for\; the\; solution.}$

$\bold{Hope\;it \; helps\;!}$

Answer 2

$\begin{gathered}\Large{\bold{\blue{\underline{Formula \: Used \::}}}} \end{gathered}$

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

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

$\rm :\implies\:\:\boxed{ \green{ \bf \: \dfrac{d}{dx} {a}^{x} = {a}^{x} log(a)}}$

$\large\underline\purple{\bold{Solution :- }}$

$\tt \longmapsto\:y = \dfrac{9}{sinx} + 3cotx + {a}^{x}$

$\tt \longmapsto\:y = 9cosecx + 3cotx + {a}^{x}$

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

$\tt \longmapsto\:\dfrac{d}{dx} y =\dfrac{d}{dx}( 9cosecx + 3cotx + {a}^{x} )$

$\tt \longmapsto\:\dfrac{dy}{dx} = 9\dfrac{d}{dx}cosecx + 3\dfrac{d}{dx}cotx + \dfrac{d}{dx}{a}^{x}$

$\tt :\implies\:\dfrac{dy}{dx}=- 9cosecx \: cotx-3 {cosec}^{2}x+{a}^{x} log(a)$