Question #54aa1

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Question #54aa1

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Answer 1 · 2017-11-15T21:55:41

$\frac{2 e^{2 x^{2}} (2 x^{2} ln x - 1)}{x {ln}^{3} (x)}$ $(2e^(2x^2)(2x^2lnx-1))/(xln^3(x))$

Explanation:

for the quotient rule

$\frac{e^{2 x^{2}}}{{(ln x)}^{2}}$ $(e^(2x^2))/(Lnx)^2$

$\frac{d}{d x} (\frac{e^{2 x^{2}}}{{(ln x)}^{2}})$ $d/dx((e^(2x^2))/(Lnx)^2)$

$\frac{d}{d x} (\frac{e^{2 x^{2}}}{{(ln x)}^{2}}) = \frac{\frac{d}{d x} (e^{2 x^{2}}) ({ln}^{2} (x)) - (e^{2 x^{2}}) \frac{d}{d x} ({ln}^{2} (x))}{{({ln}^{2} (x))}^{2}}$ $d/dx((e^(2x^2))/(Lnx)^2)=(d/dx(e^(2x^2))(ln^2(x))-(e^(2x^2))d/dx(ln^2(x)))/(ln^2(x))^2$

$\frac{d}{d x} (e^{2 x^{2}}) = 4 x . e^{2 x^{2}}$ $d/dx(e^(2x^2))=4x.e^(2x^2)$

$\frac{d}{d x} ({ln}^{2} (x)) = \frac{2}{x} ln (x)$ $d/dx(ln^2(x))=2/xln(x)$

$\frac{d}{d x} (\frac{e^{2 x^{2}}}{{(ln x)}^{2}}) = \frac{(4 x . e^{2 x^{2}}) ({ln}^{2} (x)) - (e^{2 x^{2}}) \frac{2}{x} ln (x)}{{({ln}^{2} (x))}^{2}}$ $d/dx((e^(2x^2))/(Lnx)^2)=((4x.e^(2x^2))(ln^2(x))-(e^(2x^2))2/xln(x))/(ln^2(x))^2$

multiplying everything for a x

$\frac{d}{d x} (\frac{e^{2 x^{2}}}{{(ln x)}^{2}}) = \frac{(4 x^{2} . e^{2 x^{2}}) ({ln}^{2} (x)) - (e^{2 x^{2}}) 2 ln (x)}{x {({ln}^{2} (x))}^{2}}$ $d/dx((e^(2x^2))/(Lnx)^2)=((4x^2.e^(2x^2))(ln^2(x))-(e^(2x^2))2ln(x))/(x(ln^2(x))^2$

factoring $e^{2 x^{2}} (2 ln (x))$ $e^(2x^2)(2ln(x))$

$\frac{d}{d x} (\frac{e^{2 x^{2}}}{{(ln x)}^{2}}) = \frac{2 e^{2 x^{2}} ln (x) (2 x^{2} ln x - 1)}{x {ln}^{4} (x)}$ $d/dx((e^(2x^2))/(Lnx)^2)=(2e^(2x^2)ln(x)(2x^2lnx-1))/(xln^4(x))$

dividing

$\frac{d}{d x} (\frac{e^{2 x^{2}}}{{(ln x)}^{2}}) = \frac{2 e^{2 x^{2}} (2 x^{2} ln x - 1)}{x {ln}^{3} (x)}$ $d/dx((e^(2x^2))/(Lnx)^2)=(2e^(2x^2)(2x^2lnx-1))/(xln^3(x))$

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Question #54aa1

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