Question

differentiate the following function with respect to x.

$y = {x}^{3 \div 2} \: \: {e}^{x} \: \: logx$

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Answer 1

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

Given function is

$\rm :\longmapsto\:y = {\bigg(x\bigg) }^{\dfrac{3}{2} } \: {e}^{x} \: logx$

On taking log on both sides, we get

$\rm :\longmapsto\: \: logy = log\bigg[ {\bigg(x\bigg) }^{\dfrac{3}{2} } \: {e}^{x} \: logx\bigg]$

We know,

$\red{\rm :\longmapsto\:\boxed{\tt{ logxy \: = \: logx \: + \: logy \: }}}$

So, using this identity, we get

$\rm :\longmapsto\:logy = log {\bigg(x\bigg) }^{\dfrac{3}{2} } + log {e}^{x} + log(logx)$

We know,

$\red{\rm :\longmapsto\:\boxed{\tt{ log {x}^{y} \: = \: y \: logx \: }}}$

So, using this, we get

$\rm :\longmapsto\:logy = \dfrac{3}{2}logx + xloge + log(logx)$

$\rm :\longmapsto\:logy = \dfrac{3}{2}logx + x + log(logx)$

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

$\rm :\longmapsto\:\dfrac{d}{dx}logy =\dfrac{d}{dx}\bigg[ \dfrac{3}{2}logx + x + log(logx) \bigg]$

We know,

$\red{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx}logx = \frac{1}{x} \: }}}$

So, using this, we get

$\rm :\longmapsto\: \dfrac{1}{y} \dfrac{dy}{dx} =\dfrac{3}{2}\dfrac{d}{dx}logx + \dfrac{d}{dx}x + \dfrac{d}{dx}log(logx)$

$\rm :\longmapsto\: \dfrac{1}{y} \dfrac{dy}{dx} =\dfrac{3}{2} \times \dfrac{1}{x}+ 1 + \dfrac{1}{logx}\dfrac{d}{dx}(logx)$

$\rm :\longmapsto\: \dfrac{1}{y} \dfrac{dy}{dx} =\dfrac{3}{2x} + 1 + \dfrac{1}{x \: logx}$

$\rm :\longmapsto\: \dfrac{dy}{dx} =y\bigg[\dfrac{3}{2x} + 1 + \dfrac{1}{x \: logx}\bigg]$

$\rm :\longmapsto\: \dfrac{dy}{dx} ={\bigg(x\bigg) }^{\dfrac{3}{2} } \: {e}^{x} \: logx\bigg[\dfrac{3}{2x} + 1 + \dfrac{1}{x \: logx}\bigg]$

Hence,

$\red{\boxed{\tt{ \dfrac{dy}{dx} ={\bigg(x\bigg) }^{\dfrac{3}{2} } \: {e}^{x} \: logx\bigg[\dfrac{3}{2x} + 1 + \dfrac{1}{x \: logx}\bigg]}}}$

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More to Know

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

Answer 2

Refer to the attachment.

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