# Science:Math Exam Resources/Courses/MATH110/December 2015/Question 06 (b)/Solution 1

By the product rule, we first have

Using the chain rule with , we have

Therefore, plugging this back into the above equation, we get

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By the product rule, we first have

$f'(x)=(xe^{(x^{2})})'=(x)'e^{(x^{2})}+x(e^{(x^{2})})'=e^{(x^{2})}+x(e^{(x^{2})})'.$

Using the chain rule with $u=x^{2}$, we have

$(e^{(x^{2})})'={\frac {d}{dx}}e^{(x^{2})}={\frac {d}{du}}e^{u}\cdot {\frac {du}{dx}}=e^{u}{\frac {d(x^{2})}{dx}}=e^{(x^{2})}\cdot 2x=2xe^{(x^{2})}.$

Therefore, plugging this back into the above equation, we get

${\begin{aligned}f'(x)=e^{(x^{2})}+x(e^{(x^{2})})'=e^{(x^{2})}+x\cdot (2xe^{(x^{2})})=\color {blue}(2x^{2}+1)e^{(x^{2})}.\end{aligned}}$

- This page was last edited on 14 November 2017, at 07:49.