Help with Derivatives

**roksteady** · February 20th 2009, 06:55 PM

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.

**JoshHJ** · February 20th 2009, 07:01 PM

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 $

**roksteady** · February 21st 2009, 03:29 PM

yep that's them.

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

-thankx

**JoshHJ** · February 21st 2009, 04:28 PM

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

Now to help you with your first problem:

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

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 $

**mr fantastic** · February 22nd 2009, 02:09 AM

Originally Posted by JoshHJ

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

Now to help you with your first problem:

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

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

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

**mr fantastic** · February 27th 2009, 08:01 PM

Originally Posted by roksteady

yep that's them.

[snip]

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

Originally Posted by roksteady

5. $\sqrt[3]{x}-1/\sqrt[3]{x}$
I got: 1/3x^(-2/3) + 1/3x^(-4/3)

Your answer is correct.

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

Thread: Help with Derivatives

LinkBack

Thread Tools

Display

Help with Derivatives

Similar Math Help Forum Discussions

Derivatives and Anti-Derivatives

first-order and second-order derivatives and partial derivatives

Derivatives with both a and y

Derivatives Of inverse functions; Derivatives and integrals involving exponential fun

Trig derivatives/anti-derivatives

Search Tags