Differentiate (x+1)^3(X+4)(x-3)^2

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embizze | High School Teacher | (Level 1) Educator Emeritus

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Differentiate `(x+1)^3(x+4)(x-3)^2` :

We will use the extended product rule:

`d/(dx)[f(x)g(x)h(x)]=f'(x)g(x)h(x)+f(x)g'(x)h(x)+f(x)g(x)h'(x)`

We will also use the chain rule:

`d/(dx)f(g(x))=f'(g(x))g'(x)`

So `d/(dx)[(x+1)^3(x+4)(x-3)^2]`

`=d/(dx)[(x+1)^3](x+4)(x-3)^2+(x+1)^3d/(dx)[(x+4)](x-3)^2+(x+1)^3(x+4)d/(dx)[(x-3)^2]`

`=3(x+1)^2(1)(x+4)(x-3)^2+(x+1)^3(1)(x-3)^2+(x+1)^3(x+4)(2)(x-3)(1)`

`=3(x+4)(x+1)^2(x-3)^2+(x+1)^3(x-3)^2+2(x+4)(x-3)(x+1)^3`

`=(x+1)^2(x-3)[3(x+4)(x-3)+(x+1)(x-3)+(x+1)(x+4)]`

`=(x+1)^2(x-3)[6x^2+11x-31]`

`=(x+1)^2[6x^3-7x^2-64x+93]`

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The derivative is `(x+1)^2[6x^3-7x^2-64x+93]`

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** If you get rid of all parentheses you will get:

`f'(x)=6x^5+5x^4-72x^3-42x^2+122x+93`

If you expand the given product you will get:

`f(x)=x^6+x^5-18x^4-14x^3+61x^2+93x+36`

Differentiating term by term using the general power rule will yield the same derivative.

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