# Deriving a root and a fraction

1. Oct 16, 2011

### Twinflower

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

Derive $y= \sqrt(2x^4) - \frac{5}{3x^2}$

3. The attempt at a solution

I am still at the first part of the function (the root):
First I tried to derive inside the root like this:
$$\sqrt(2x^4) = \sqrt(8x^3) = 2 \sqrt(2) \times \sqrt(x^3) = \sqrt(2) \times x^2$$

Unfortunately, the first part of the function is supposed to be 2\sqrt(2)x.

The second part I am nowhere close yet.

2. Oct 16, 2011

### I like Serena

Hi Twinflower!

Before you take the derivative, you should first try to simplify your expression.

Did you know that $\sqrt {(2x)} = \sqrt 2 \sqrt x$?
And that $\sqrt {(x^4)} = x^2$?

Last edited: Oct 16, 2011
3. Oct 16, 2011

### Twinflower

I tried what you are suggesting, but I somehow ended up with $\sqrt(x^4) = x^3$ since $\sqrt(x^2) = x^1 = x$

Second attempt:

$$\sqrt(2x^4)$$
$$\sqrt(2) \times \sqrt(x^4)$$
$$\sqrt(2) \times x^2$$
$$2\times \sqrt(2) \times x$$
$$2 x$$

hm.... didn't go quite well now either?

4. Oct 16, 2011

### I like Serena

Actually, your second attempt is looking quite well. :)

But how did you get from:

$2\times \sqrt(2) \times x$

to

$2 x$?

5. Oct 16, 2011

### Twinflower

Brain-fart.

For a second I though that 2 times sqrt2 equals 2.

The final answer should ofcourse be
$$2 \times \sqrt(2) \times x$$

6. Oct 16, 2011

### Twinflower

Thank you by the way.

Now I need a push in the right direction regarding the fraction :)

7. Oct 16, 2011

### I like Serena

You're welcome!

For the fraction you need to know that for instance ${1 \over x^5} = x^{-5}$.

Oh, and also that ${2 \over 3x} = {2\over 3} \times {1 \over x}$.

8. Oct 16, 2011

### Twinflower

Ah, of course!
Gimme a minute, and I'll figure that one out as well :)

9. Oct 16, 2011

### Twinflower

Ok, here we go:

$$- \frac{5}{3x^2}$$
$$- \frac{5}{3} \times \frac{1}{x^2}$$
$$- \frac{5}{3} \times x^{-2}$$
$$- \frac{5}{3} \times -2 \times x^{-3}$$
$$\frac{10}{3} \times x^{-3}$$
$$\frac{10\times x^{-3}}{3}$$

10. Oct 16, 2011

### Twinflower

Thanks a bunch! :)

11. Oct 16, 2011