How do you differentiate $g (x) = e^{x} \cdot \frac{1}{x^{2}}$ $g(x) =e^x*1/x^2$ using the product rule?

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Answer 1 · 2016-05-26T17:55:50

$\frac{e^{x}}{x^{2}} - \frac{2 e^{x}}{x^{3}}$ $\frac{e^x}{x^2}-\frac{2e^x}{x^3}$

$\frac{d}{d x} (e^{x} \frac{1}{x^{2}})$ $\frac{d}{dx}(e^x\frac{1}{x^2})$

Applying product rule, $(fcdot g)^'=f^'cdot g+fcdot g^'$

$f=e^x,g=\frac{1}{x^2}$

$=\frac{d}{dx}(e^x)\frac{1}{x^2}+\frac{d}{dx}(\frac{1}{x^2})e^x$

We know,
$\frac{d}{dx}(e^x)=e^x$ and

$\frac{d}{dx}(\frac{1}{x^2})=-\frac{2}{x^3}$

So,
$=e^x\frac{1}{x^2}+(-\frac{2}{x^3})e^x$

Finally, simplifying it,
$=\frac{e^x}{x^2}-\frac{2e^x}{x^3}$

How do you differentiate g(x) =e^x*1/x^2g(x)=ex⋅1x2g(x) =e^x*1/x^2 using the product rule?