1. ## Help 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.

2. Before I work this through, tell me if this is what the equation is:
$
x^3 - \sqrt[3]{x}
$

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

3. yep that's them.

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

-thankx

4. Just use this tutorial: http://www.mathhelpforum.com/math-he...-tutorial.html

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

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

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

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

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

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

5. Originally Posted by JoshHJ
Just use this tutorial: http://www.mathhelpforum.com/math-he...-tutorial.html

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

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

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

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

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

$f'(x) = (s^2 + s + 1)^4 * 7(s-2)^6 + (s-2)^7 * 8s(s^2 + s + 1)^3$
I think you might have meant $f({\color{red}s}) = (s^2 + s + 1)^4 * (s-2)^7$

and

$f'({\color{red}s}) = (s^2 + s + 1)^4 * 7(s-2)^6 + (s-2)^7 * {\color{red}4(2s+1)} (s^2 + s + 1)^3$.

yep that's them.

[snip]
Clearly not, since you re-posted the cube root one ....

5. $\sqrt[3]{x}-1/\sqrt[3]{x}$