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

1 Answer

Shwetank Mauria

Jun 18, 2016

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

Explanation:

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)#

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