How do you differentiate $y=e^-x/x$ ?

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Answer 1 · 2016-10-21T21:19:49

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

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

so, $dy/dx = (xd/dx(e^-x)-e^-xd/dx(x)) / x^2$

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

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

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

How do you differentiate y=e^-x/xy=e^-x/x?