What is the derivative of $f (x) = \frac{e^{\frac{1}{x}}}{x^{2}}$ $f(x)=(e^(1/x))/x^2$ ?

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Answer 1 · 2014-09-20T16:22:19

$f'(x)={-1/x^2e^{1/x}cdot x^2-e^{1/x}cdot2x}/{(x^2)^2}$

by cancelling out $x^2$ in the numerator,

$={-e^{1/x}-2xe^{1/x}}/{x^4}$

by factoring out $-e^{1/x}$ from the numerator,

$=-{(1-2x)e^{1/x}}/{x^4}$

What is the derivative of f(x)=(e^(1/x))/x^2f(x)=e1xx2f(x)=(e^(1/x))/x^2 ?