# What is the derivative of this function?f(x)= (2x+1)^3(3x-x^2)

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

Break it in the middle for a product rule:

d(first)*second + d(second)*first

To find derivative of the first piece, `(2x+1)^3` , we use chain rule:

d(inside)*d(outside with inside the same)

=`2*3(2x+1)^2`

=`6(2x+1)^2`

Derivative of the second piece, `3x-x^2` , is just `3-2x` .

Put it all together:

d(first)*second + d(second)*first =

`6(2x+1)^2*(3x-x^2)` + `(3-2x)*(2x+1)^3`

Maybe you could factor the (2x+1)^2 out:

`(2x+1)^2*(6(3x-x^2)+(3-2x)(2x+1))`

That big second factor is only 2nd-order, so let's simplify that bad boy:

`(2x+1)^2(18x-6x^2+6x+3-4x^2-2x)`

`(2x+1)^2(-10x^2+22x+3)`

For even better looks, factor a -1 out of that second factor:

`-(2x+1)^2(10x^2-22x-3)`

Cha-ching! Let me know what you think.