How do you differentiate #f(x)=8x(2x+1)^2 + 3(2x+1)^3# using the product rule?

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Answer 1 · 2015-12-06T16:21:27

#f'(x)=168x^2+136x+26#

Find the derivative of each part.

#d/dx[8x(2x+1)^2]=(2x+1)^2d/dx[8x]+8xd/dx[(2x+1)^2]#

#=8(2x+1)^2+16x(2x+1)d/dx[2x+1]#

#=8(2x+1)^2+32x(2x+1)#

#=96x^2+64x+8#

#d/dx[3(2x+1)^3]=9(2x+1)^2d/dx[2x+1]#

#=18(2x+1)^2#

#=72x^2+72x+18#

Add the two derivatives we've found:

#f'(x)=96x^2+64x+8+72x^2+72x+18#

#=168x^2+136x+26#