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

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Answer 1 · 2016-08-18T15:42:03

#dy/dx=7x^6-6x^5+30x^4-24x^3+ 36x^2-24x+8#

Let #u=(x-1) -> (du)/dx=1#

Let #v=(x^2+2)^3->(du)/dx=3(x^2+2)^2(2x) =6x(x^2+2)^2#

Set #y=(x-1)(x^2+2)^3 ->uv#

Using #dy/dx=v(du)/dx+u(dv)/dx#
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#dy/dx" "=" "[(x^2+2)^3xx1] + (x-1)[6x(x^2+2)^2] #......Eqn(1)

#=>dy/dx" "=" "(x^2+2)^2[(x^2+2)+6x(x-1)]#

#=>dy/dx" "=" "(x^2+2)^2(7x^2-6x+2)#

#=>dy/dx" "=" "(x^4+4x^2+4)(7x^2-6x+2)#

#dy/dx=color(blue)(7x^6 -6x^5+2x^4)color(green)(+28x^4-24x^3+8x^2)color(purple)(+28x^2-24x+8)#

#dy/dx=7x^6-6x^5+30x^4-24x^3+ 36x^2-24x+8#

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Tony B