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How do you differentiate #y=e^x/x^2#?

1 Answer

Nov 30, 2016

The answer is #=(e^x(x-2))/x^3#

Explanation:

We use the derivative of a quotient

#(u/v)'=(u'v-uv')/(v^2)#

#u=e^x#, #=>#, #u'=e^x#

#v=x^2#, #=>#, #v'=2x#

So, #dy/dx=(e^x*x^2-e^x*2x)/(x^4)#

#=(xe^x(x-2))/(x^4)=(e^x(x-2))/x^3#

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