I need some help solving these derivative problems:

1. F(x)= 3rd root of(x)-[1/3rd root of(x)]

For this one I got: 1/3 x - 3rd root of(9)

2. (s^2+s+1)^4 * (s-2)^7

I tried doing this one but it got out of hand and messy.

Any help would be great.

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- Feb 20th 2009, 06:55 PMroksteadyHelp with Derivatives
I need some help solving these derivative problems:

1. F(x)= 3rd root of(x)-[1/3rd root of(x)]

For this one I got: 1/3 x - 3rd root of(9)

2. (s^2+s+1)^4 * (s-2)^7

I tried doing this one but it got out of hand and messy.

Any help would be great. - Feb 20th 2009, 07:01 PMJoshHJ
Before I work this through, tell me if this is what the equation is:

$\displaystyle

x^3 - \sqrt[3]{x}

$

and

$\displaystyle

(s^2 + s + 1)^4 * (s-2)^7

$ - Feb 21st 2009, 03:29 PMroksteady
yep that's them.

For future reference how do you do that (make math symbols visible and all)

-thankx - Feb 21st 2009, 04:28 PMJoshHJ
Just use this tutorial: http://www.mathhelpforum.com/math-he...-tutorial.html

Now to help you with your first problem:

$\displaystyle f(x) = x^3 - {x}^\frac {1}{3}$

From here, you will want to use the power rule so:

$\displaystyle f'(x) = 3x^2 - \frac {x}{3}^ {\frac {1}{3} - 1} = $$\displaystyle 3x^2 - \frac {x}{3}^ {\frac {-2}{3}}$

Now for the second you will need to use the chain rule and the product rule.

$\displaystyle

f(x) = (s^2 + s + 1)^4 * (s-2)^7

$

$\displaystyle

f'(x) = (s^2 + s + 1)^4 * 7(s-2)^6 + (s-2)^7 * 8s(s^2 + s + 1)^3

$

- Feb 22nd 2009, 02:09 AMmr fantastic
- Feb 27th 2009, 08:01 PMmr fantastic
Clearly not, since you re-posted the cube root one ....

Quote:

Originally Posted by**roksteady**

Note: The derivative of the first term is done as has already been shown in this thread.

However, I'm glad to see you learned some latex (Clapping)