Question

${\boxed{ \bold \red{ \frac{d}{dx} \bigg( \cos( \sqrt{x} ) \bigg)}}}$


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Answer 1

$\frac{d}{dx} ( \cos (\sqrt{x} ))$

$- \sin( \sqrt{x} ) \cdot \frac{d}{dx} ( \sqrt{x})$

$- \sin (\sqrt{x}) \frac{d}{dx} (x {}^{ \frac{1}{2} } )$

$- \sin( \sqrt{x} ) \frac{1}{2} x {}^{ \frac{ - 1}{2} }$

$\frac{ - \sin( \sqrt{x} ) }{2 \sqrt{x} }$

Answer 2

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

Given function is

$\red{\rm :\longmapsto\:cos \sqrt{x}}$

Let assume that

$\rm :\longmapsto\:y = cos \sqrt{x}$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx}cos \sqrt{x}$

We know,

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

So, using this identity, we get

$\rm :\longmapsto\:\dfrac{dy}{dx} = - \: sin \sqrt{x} \: \dfrac{d}{dx} \sqrt{x}$

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

We know,

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} \: }}}$

So, using this identity, we get

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

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

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

$\rm \implies\:\boxed{ \tt{ \: \dfrac{dy}{dx} = - \: \frac{sin \sqrt{x} }{2 \sqrt{x} } \: }}$

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Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$