A student asks:

How do I simplify horrible product and quotient rule expressions?

The example they gave was differentiating:

$f(x) = (2x - 3)^4 (x^2 + x + 1)^5$

First up, a careful bit of product rule:

$u = (2x - 3)^4$, so $\diff{u}{x} = 8(2x- 3)^3$

$v = (x^2 + x + 1)^5$, so $\diff{v}{x} = 5(2x+1)(x^2 + x + 1)^4$

... and we want $u \diff{v}{x} + v \diff{u}{x}$.

Well now. I know students who would methodically multiply out the brackets using the binomial expansion, end up with an awful polynomial of degree 13 and say "what now?"

Don't do that. The trick is to spot common factors almost before you start: $u$ and $\diff ux$ share a factor of $(2x - 3)^3$. $v$ and $\diff vx$ have $(x^2 + x + 1)^4$ in common. We should take those out before we do anything else.

$u \diff vx = (2x - 3)^3 (x^2 + x + 1)^4 [5(2x- 3) (2x+1)]$

$v \diff ux = (2x - 3)^3 (x^2 + x + 1)^4 [8(x^2 + x + 1)]$

Now we need to expand the square brackets before we add them: the first is $20x^2 - 20x - 15$ and the second is $8x^2 + 8x + 8$. Adding them up gives $28x^2 - 12x -7$ (which doesn't factorise), so our final answer is:

$ (2x - 3)^3 (x^2 + x + 1)^4 [ 28x^2 -12x - 7]$.

It's worth making sure the lead term is correct as a sanity check: if you expanded the original expression, you'd start with $(2x)^4(x^2)^5 = 16x^{14}$ - so you're looking for $224x^{13}$ as the first term of the final Thing: our lead term in the derivative is $(2x)^3 (x^2)^4 (28x^2) = (8)(28)x^{13} = 224x^{13}$. Phew!

*Edited 2015-01-26 to correct a bracket*

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008.
He lives with an espresso pot and nothing to prove.

## RealityMinus3

@icecolbeveridge Oooh v tidy.

## MathbloggingAll

Simplifying http://t.co/EBek4m05Lf

## MarkkuOpe

Derivointi ja yhteinen tekijä, mutta muista tarkistaa, vaikkapa vain korkeimman asteen kerroin. #UrMaa6, #UrMaa13

http://t.co/1F8hqeW9Tn