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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Check:
Tony B

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