# Differentiation problem

1. Jan 14, 2009

### breen155

Hey I'm stuck on a problem, i have to find the differential of y=5x(2x-1)^3 and therefore find the x-coordinate of a stationary point on a graph,
I use the chain rule and get the differential to be 30x(2x-1)^2 and therefore the x-coordinate to be 0.5
however my textbook says the differential is 5(2x-1)^2(8x-1) and the x-coordinate of the stationary point is 1/8th I was wondering if someone could explain how this answer was obtained. Thanks

2. Jan 14, 2009

### CompuChip

You used the chain rule, but you forgot about the product rule?
If you open up the last pair of brackets the first term is precisely what you got.

3. Jan 14, 2009

### breen155

Ah thank you very much but i was always taught to do the chain rule in equations like the one above but the product rule worked :P again, thanks.

4. Jan 14, 2009

### HallsofIvy

Staff Emeritus
Sometimes you need both!

5. Jan 14, 2009

### CompuChip

If you have a product of two (or more things) you always start with the product rule,
(f(x) g(x))' = f'(x) g(x) + f(x) g'(x)

This obviously requires you to calculate f'(x) and g'(x) and you may need the chain rule for either f'(x), or g'(x), or both. For example, for the derivative of
(3x + 6)^2 (2x - 1)^3
you will need the product rule once and the chain rule twice.

6. Jan 14, 2009

### breen155

I understand now. Thanks for all the help guys :)