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How do you differentiate #g(y) =(x^2 - 2x + 1)^4 (4x^6 + 5) # using the product rule?

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Ben Share
Jun 13, 2018

Answer:

Treat the two terms as a normal product rule, but then use the chain rule on the first function when differentiating to get the full solution: #g'(x)=8(x-1)^7(7x^6-3x^5+5)#

Explanation:

First, let's designate the two pieces of the function as #f_1(x)# and #f_2(x)#:

#f_1(x)=(x^2-2x+1)^4#
#f_2(x)=4x^6+5#

For a given function, the derivative using the product rule is:

#(dy)/(dx)(f_1(x)f_2(x))=f_1'(x)*f_2(x)+f_1(x)*f_2'(x)#

This means we'll need to know the derivatives of each #f# function. The first function requires a chain rule expansion:

#(dy)/(dx)f(g(x))=f'(g(x))*g'(x)#

Deriving #f_1'(x)#:

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

#f_1'(x)=(x^2-2x+1)^3(8x-8)#

Deriving #f_2'(x)#:

#(dy)/(dx)(4x^6+5)=24x^5#

Now, we reassemble (and simplify):

#g'(x)=(x^2-2x+1)^3(8x-8)(4x^6+5)+24x^5(x^2-2x+1)^4#

#g'(x)=8((x^2-2x+1)^3(x-1)(4x^6+5)+3x^5(x^2-2x+1)^4)#

Note that the factored form of #x^2-2x+1# is #(x-1)^2#

#g'(x)=8(((x-1)^2)^3(x-1)(4x^6+5)+3x^5((x-1)^2)^4)#

#g'(x)=8((x-1)^6(x-1)(4x^6+5)+3x^5(x-1)^8)#

#g'(x)=8((x-1)^7(4x^6+5)+3x^5(x-1)^8)#

#g'(x)=8(x-1)^7((4x^6+5)+3x^5(x-1))#

#g'(x)=8(x-1)^7(4x^6+5+3x^6-3x^5)#

#color(green)(g'(x)=8(x-1)^7(7x^6-3x^5+5)#

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