Question #1092f

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Answer 1 · 2015-02-23T12:35:51

The answer is: #h'(x)=(2x+7)^3(x-1)(14x^2+11x-7)#.

#h'(-3)=344#

Rememering that:

#y=f(x)*g(x)rArry'=f'(x)*g(x)+f(x)g'(x)#

Since:

#h(x)=x(2x+7)^4(x-1)^2#

than:

#h'(x)=1*(2x+7)^4*(x-1)^2+x*4(2x+7)^3*2*(x-1)^2+x(2x+7)^4*2(x-1)^1*1=#

#=(2x+7)^3(x-1)*[(2x+7)(x-1)+8x(x-1)+2x(2x+7)]=#

#=(2x+7)^3(x-1)*(2x^2-2x+7x-7+8x^2-8x+4x^2+14x)=#

#=(2x+7)^3(x-1)(14x^2+11x-7)#.

#h'(-3)=(2*(-3)+7)^3(-3-1)(14*(-3)^2+11*(-3)-7)=#

#=(1)^3(-4)(86)=344#