How do you differentiate #g(y) =(y^-1 + y^-2)(5y^2 -y^3)# using the product rule?

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Answer 1 · 2016-07-02T19:04:02

#g'(y)=4-2y#.

#g(y)=(y^-1+y^-2)(5y^2-y^3)=(y^-1+y^-2)y^2(5-y)=(y^-1*y^2+y^-2*y^2)(5-y)=(y+1)(5-y)#

Using Product Rule #: d/dy(U*V)=U*d/dyV+V*d/dyU#

#g'(y)=(y+1)d/dy(5-y)+(5-y)d/dy(y+1)#

#=(y+1){d/dy5-d/dyy}+(5-y){d/dyy+d/dy1}#

#=(y+1)(0-1)+(5-y)(1+0)=-y-1+5-y=4-2y#

Note that we were reqd. to diif. using Product Rule. Otherwise,

#g(y)=(y+1)(5-y)=5y+5-y^2-y=5+4y-y^2#' so that,

#g'(y)=(5)'+(4y)'-(y^2)'=0+4*1-2y=4-2y#, as before!

I hope this will be of help! Enjoy Maths.!