How do you differentiate # y = ((2x - 1)^3)((x + 1)^3)#?

1 Answer
Shwetank Mauria
Sep 14, 2016

#(dy)/(dx)=3(2x-1)^2(x+1)^2(4x+1)#

Explanation:

Product rule states if #y(x)=g(x)h(x)#

then #(dy)/(dx)=(dg)/(dx)xxh(x)+(dh)/(dx)xxg(x)#

Hence as #y(x)=(2x-1)^3(x+1)^3#

#(dy)/(dx)=3xx(2x-1)^2xx2xx(x+1)^3+3xx(x+1)^2xx1xx(2x-1)^3#

Here we have also used the concept of function of a function and used chain rule for it. As we have differentiated #(2x-1)^3# and #(x+1)^3# w.r.t. #(2x-1)# and #(x+1)#, we must multiply by differential of #(2x-1)# and #(x+1)# w.r.t #x# i.e. by #2# and #1# respectively

So #(dy)/(dx)=6(2x-1)^2(x+1)^3+3(x+1)^2(2x-1)^3#

= #3(2x-1)^2(x+1)^2[2(x+1)+2x-1]#

= #3(2x-1)^2(x+1)^2[2x+2+2x-1]#

= #3(2x-1)^2(x+1)^2(4x+1)#

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