How do you differentiate $y = \frac{e^{- x}}{x}$ $y=e^-x/x$ ?

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How do you differentiate $y = \frac{e^{- x}}{x}$ $y=e^-x/x$ ?

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Answer 1 · 2016-10-21T21:19:49

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

Explanation:

You need to use the quotient rule; $\frac{d}{d x} (\frac{u}{v}) = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v^{2}}$ $d/dx(u/v) = (v(du)/dx-u(dv)/dx)/v^2$

so, $\frac{d y}{d x} = \frac{x \frac{d}{d x} (e^{- x}) - e^{- x} \frac{d}{d x} (x)}{x^{2}}$ $dy/dx = (xd/dx(e^-x)-e^-xd/dx(x)) / x^2$

$:. dy/dx = (x(-e^-x)-e^-x(1)) / x^2$

$:. dy/dx = (-xe^-x-e^-x) / x^2$

$:. dy/dx = (-e^-x(x+1)) / x^2$

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How do you differentiate $y = \frac{e^{- x}}{x}$ $y=e^-x/x$ ?

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