How do you differentiate $f(x)=e^-x/x$ ?

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Answer 1 · 2017-03-10T17:34:36

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

According to Quotient rule if $f(x)=(g(x))/(h(x))$

then $(df)/(dx)=((dg)/(dx)xxh(x)-(dh)/(dx)xxg(x))/(h(x))^2$

As $f(x)=e^(-x)/x$

$(df)/(dx)=(-e^(-x)xx x-1xxe^(-x))/x^2$

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

How do you differentiate f(x)=e^-x/xf(x)=e^-x/x?