# Fourth Derivitive Problem ~ (Need Help for Calc Midterm)

• Jan 14th 2009, 06:08 PM
Power Rules
Fourth Derivitive Problem ~ (Need Help for Calc Midterm)
I've got a question that asks me to take the 4th derivitive of (2x+1)^4 ~ The problem I come across is once it becomes 24(2x+1)^0 I don't understand what the value is...

I'm not sure if I'm correct on this but (2x+1)^4 differentiated is:
4(2x+1)^3
12(2x+1)^2
24(2x+1)^1
24(2x+1)^0

Can anyone tell me if I did something wrong, and then answer my question? >.O
• Jan 14th 2009, 06:10 PM
Jameson
Hey there. You forgot to use the chain rule for all your derivatives so they will all be off.

d/dx (2x+1)^4 = 4(2x+1)^3 * 2
• Jan 14th 2009, 06:13 PM
Power Rules
Alright, someone brought the Chain Rule up to me earlier when I asked also

So, I thought that worked but once I take the first derivitive I became confused because 4(2x+1)^3 *2 when differentiated again for the second derivitive would get rid of the *2 wouldn't it? Or am I completely wrong?
• Jan 14th 2009, 06:15 PM
Jameson
Quote:

Originally Posted by Power Rules
Alright, someone brought the Chain Rule up to me earlier when I asked also

So, I thought that worked but once I take the first derivitive I became confused because 4(2x+1)^3 *2 when differentiated again for the second derivitive would get rid of the *2 wouldn't it? Or am I completely wrong?

Before differentiating again, simplify what you have to make it easier.

$\displaystyle 4(2x+1)^3 \times 2 = 8(2x+1)^3$

Now repeat process - $\displaystyle \frac{d}{dx} 8(2x+1)^3 = 24(2x+1)^2 \times 2$ See the pattern?
• Jan 14th 2009, 06:23 PM
Power Rules
Alright, that helps a lot. So if I simplify it essentially keeps me from any confusion about what stays and what goes, because the coefficient is all that changes due to the chain rule. Thank you, that really does help.

If only I'd found this place before the test I had today, I might have done a little better on it. (We're on integration, confusion about the chain rule should have been repaired long ago)
• Jan 14th 2009, 06:29 PM
Jameson
Well welcome to the forums and I hope you stick around for all your math help needs :)