# 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!

Why do you think it is not correct?

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

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 ...

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

When you say it I think so, yes.

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

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

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?

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