Question

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

Answer 1

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

Explanation:

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)`