Question

differentiate wrt x :- 2x^3e^x + sqrt.x logx​

Answer 1

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

Given function is

$\rm :\longmapsto\:y = {2x}^{3} {e}^{x} + \sqrt{x} \: logx$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx} {2x}^{3} {e}^{x} + \dfrac{d}{dx} \sqrt{x} \: logx$

We know,

$\red{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx}uv = u\dfrac{d}{dx}v \: + \: v\dfrac{d}{dx}u \: }}}$

So, using this identity, we get

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

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

$\rm \:  =  \: 2 {x}^{2} {e}^{x} \bigg[ x + 3 \bigg] + \bigg[\dfrac{1}{ \sqrt{x} } + \dfrac{logx}{2 \sqrt{x} } \bigg]$

Hence,

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

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Additional Information :-

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

Answer 2

Question:-

$\sf y = 2x {}^{3} e {}^{x} + \sqrt{x} \: logx$

Given:-

$\sf y = 2x {}^{3} e {}^{x} + \sqrt{x} \: logx$

Solution:-

We know that,

$\sf \pink{ \frac{d}{dx} uv = u \frac{d}{dx} v + v \frac{d}{dx} u}$

$\sf \longmapsto \frac{dy}{dx} = 2 \bigg[x {}^{3} \frac{d}{dx} e {}^{x} + e {}^{x} \frac{d}{dx} x {}^{3} \bigg ] + \bigg[ \sqrt{x} \frac{d}{dx} log x + logx \frac{d}{dx} \sqrt{x} \bigg]$

$\sf = 2 \bigg[x {}^{3} e {}^{x} + e {}^{x} (3x {}^{2}) \bigg ] + \bigg[ \sqrt{x} \times \frac{1}{x} + logx \times \frac{1}{2 \sqrt{x} } \bigg ]$

$\sf = 2x {}^{2} e {}^{x} \bigg[x + 3 \bigg] + \bigg[ \frac{1}{ \sqrt{x} } + \frac{logx}{2 \sqrt{x} } \bigg ]$

Answer:-

Hence, Required Answer is

$\sf \green{ \frac{dy}{dx} = 2x {}^{2} e {}^{x} \bigg[x + 3 \bigg] + \bigg[ \frac{1}{ \sqrt{x} } + \frac{logx}{2 \sqrt{x} } \bigg ]}$

Hope you have satisfied.