# Derivation with x^2 and root x

## 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
Staff Emeritus
Gold Member
Why do you think it is not correct?

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

robphy
Homework Helper
Gold Member
Can you simplify your last expression?

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

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
Homework Helper
Gold Member
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?

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