Derivation with x^2 and root x

[disregardthat]

Homework Statement

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

Homework Equations

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

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!

 

[Hurkyl]

Why do you think it is not correct?

 

[radou]

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

 

[robphy]

Can you simplify your last expression?

 

[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 ...

 

[Hurkyl]

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

 

[disregardthat]

When you say it I think so, yes.

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

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

 

[robphy]

When you say it I think so, yes.

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

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

missing exponent?

 

[disregardthat]

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