# Homework Help: Find the derivative of the expression and simplify fully.

1. Jul 15, 2011

### 1irishman

1. The problem statement, all variables and given/known data

k(x) = x^2(2x^3+x)^1/3

2. Relevant equations
I used the product rule and the chain rule to get to the point that i did.

3. The attempt at a solution

dy/dx(x^2)(2x^3+x)^1/3 + dy/dx(2x^3+x)^1/3(x^2)
= (2x)(2x^3+x)^1/3 + 1/3(2x^3+x)^-2/3(6x^2+1)(x^2) I can't figure out how to solve further.

2. Jul 15, 2011

### timthereaper

A couple of questions:

1) You state k(x) =... and then end up with dy/dx. Is it supposed to be y(x) at the beginning?

2) Why are you doing this: "dy/dx(x^2)(2x^3+x)^1/3 + dy/dx(2x^3+x)^1/3(x^2)"? There's no reason to add them together.

Your solution is correct as far as I've been able to check out. This is how I would rewrite what you have there: dy/dx = (2x)(2x^3+x)^1/3 + 1/3(2x^3+x)^-2/3(6x^2+1)(x^2)

You can also check your solution with wolframalpha.com.

3. Jul 15, 2011

### 1irishman

The solution is in my book, but i just don't understand how they got it.

4. Jul 15, 2011

### 1MileCrash

$Dx[x^2(2x^3+x)^\frac{1}{3}]$
=

$(x^2)Dx[(2x^2+x)^\frac{1}{3}] + (2x)(2x^3+x)^\frac{1}{3}$

..

$Dx[(2x^2+x)^\frac{1}{3}]$

let 1(x) = $x^\frac{1}{3}$
let 2(x) = u = (2x^3+x)

$Dx[mess] = 1'(u)2'(x)$
=
$\frac{1}{3}(2x^3+x)^-{\frac{2}{3}}(6x^2+1)$

..
$Dx[x^2(2x^3+x)^\frac{1}{3}]$
=
$(x^2)\frac{1}{3}(2x^3+x)^-\frac{2}{3}(6x^2+1) + (2x)(2x^3+x)^\frac{1}{3}$

=

$(\frac{x^2(6x+1)}{3(2x^3+x)^\frac{2}{3}})+(2x)(2x^3+x)^\frac{1}{3}$

That's just how I'd write it, fewer terms and no negative exponents is my preference.

5. Jul 15, 2011

### 1irishman

in your last line, is that not supposed to be (6x^2 + 1)?

6. Jul 15, 2011

### 1MileCrash

Yes! Sorry.

7. Jul 15, 2011

### Staff: Mentor

Not mentioned is that you are using dy/dx improperly. dy/dx is the derivative of y with respect to x. To indicate that you are about to differentiate with respect to x, use the operator d/dx.

Here is the corrected version of what you show above.

k'(x) = d/dx[(x^2)(2x^3+x)^(1/3)] + [STRIKE]dy/dx(2x^3+x)^1/3(x^2)[/STRIKE]
= (2x)(2x^3+x)^(1/3) + 1/3(2x^3+x)^(-2/3)(6x^2+1)(x^2)

There is some additional work that can be done to put this in a more useful form, using factoring, but all the hard work (i.e., calculus) has been done.

8. Jul 15, 2011

### 1irishman

okay, i came to a fully simplified solution starting right after the above line, here it is:

= 2x(2x^3+x)^1/3 + 1/3(1/(2x^3+x)^2/3(6x^2+1)(x^2)
= 2x(2x^3+x)^1/3 + 6x^4 + x^2/3(2x^3 + x)^2/3
= 3(2x^3 + x)^2/3 (2x)(2x^3 + x)^1/3/3(2x^3 + x)^2/3 + 6x^4 + x^2/3(2x^3 + x)^2/3
= 6x(2x^3 + x) + 6x^4 + x^2/3(2x^3+ x)^2/3
= 12x^4 + 6x^2 + 6x^4 + x^2/3(2x^3 + x)^2/3
= 18x^4 + 7x^2/3(2x^3+x)^2/3
= 1/3x^2(18x^2 + 7)(2x^3 + x)^-2/3
It said to simplify fully so that is what i did.

9. Jul 15, 2011

### 1irishman

thank you.

10. Jul 15, 2011

### Staff: Mentor

You're welcome!