Question

d/dx sin^5 (underoot x)​

Answer 1

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

$\rm :\longmapsto\:Let \: y \: = \: {sin}^{5} \sqrt{x}$

$\rm :\longmapsto\:y = {\bigg(sin \sqrt{x} \bigg) }^{5}$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx} {\bigg(sin \sqrt{x} \bigg) }^{5}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = 5 {\bigg(sin \sqrt{x} \bigg) }^{4} \dfrac{d}{dx} \: sin\sqrt{x}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} \bigg \}}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = 5 {\bigg(sin \sqrt{x} \bigg) }^{4} \: cos\sqrt{x} \: \dfrac{d}{dx} \sqrt{x}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \: \dfrac{d}{dx} \: sinx = \: cosx \bigg \}}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = 5 {\bigg(sin \sqrt{x} \bigg) }^{4} \: cos \sqrt{x} \: \dfrac{1}{2 \sqrt{x} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{ \because \dfrac{d}{dx} \sqrt{x} \: = \: \dfrac{1}{2 \sqrt{x} } \bigg \}}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{5 {sin}^{4} \sqrt{x} \: cos \sqrt{x} }{2 \sqrt{x} }$

Additional Information :-

$\red{\rm :\longmapsto\:\dfrac{d}{dx}k = 0}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx}x = 1}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx}cosx = - \: sinx}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx}cosecx = - \: cosecx \: cotx}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx}secx = \: secx \: tanx}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx}tanx = \: sec ^{2} x \:}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx}cotx = - \: cosec ^{2} x \:}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx} {e}^{x} = {e}^{x}}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx} {a}^{x} = {a}^{x} \: loga}$