# Derivation with x^2 and root x

1. Mar 19, 2007

### disregardthat

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

derivate: $$x^2 \sqrt{x}$$

2. Relevant equations

$$(u \cdot v)^\prime = u^\prime \cdot v + v^\prime \cdot u$$

3. The attempt at a solution

I found:
$$(x^2)^\prime = 2x$$
$$(\sqrt{x})^\prime = \frac{1}{2 \sqrt{x}}$$

I entered them into the equation:

$$2x \cdot \sqrt{x} + x^2 \cdot \frac{1}{2 \sqrt{x}} = 2x \sqrt{x} + \frac{x^2}{2 \sqrt{x}}$$

But this is not correct!

2. Mar 19, 2007

### Hurkyl

Staff Emeritus
Why do you think it is not correct?

3. Mar 19, 2007

Try to get rid of the square root in the denominator.

4. Mar 19, 2007

### robphy

Can you simplify your last expression?

5. Mar 19, 2007

### disregardthat

Ok, sorry. I checked it on my calculator many times and never got the correct result. But this time I did, by adding some extra brackets.. My calc just won't accept things with having it stuffed in with brackets!

Simplify? I guess you could simplify it:
$$2x \sqrt{x} + \frac{x^2}{2 \sqrt{x}} = \frac{4x^{1.5}}{2} + \frac{x^{1.5}}{2} = \frac{5x^{1.5}}{2} = \frac{5x \sqrt{x}}{2}$$

Jesus christ, now my textbook's answer is correct too!
Well, I guess that's a good thing. I must have seen on it uncorrectly. I got it right now... Much making a topic about ...

6. Mar 19, 2007

### Hurkyl

Staff Emeritus
FYI, you can do this problem without the product rule -- do you see how?

7. Mar 19, 2007

### disregardthat

When you say it I think so, yes.

$$x^2 \sqrt{x} = x^{2+0.5} = x^{2.5} = x^{\frac{5}{2}}$$

$$(x^{\frac{5}{2}})^\prime = \frac{5}{2} \cdot x = \frac{5x}{2}$$

8. Mar 19, 2007

### robphy

missing exponent?

9. Mar 20, 2007

### disregardthat

Of course
$$\frac{5x^{\frac{3}{2}}}{2}$$