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

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Answer 1 · 2016-06-18T05:39:24

$(df)/(dy)=e^y((y-1)/y^2)$

We use quotient rule here. According to quotient rule

if $f(x)=(g(x))/(h(x))$

$(df)/(dx)=(h(x)*g'(x)-g(x)*h'(x))/(h(x))^2$

As $f(y)=e^y/y$

$(df)/(dy)=(yxxe^y-e^yxx1)/y^2$

= $e^y((y-1)/y^2)$

How do you differentiate f(y) = e^y /yf(y) = e^y /y?