# Differentiation problem! square root function

1. Oct 20, 2011

### CrossFit415

I'm on mobile so I can't use latex.

Differentiate:

g(x)=√(4-x^4) , x is a set of [-√2, √2] and determine the domains.

So I got the derivative which is,

g'(x)=(-4x^3) ^1/2

What should I do with the -+√2 ?

I don't know what to do next

2. Oct 20, 2011

### Staff: Mentor

Nope. You need to use the chain rule. It will help to write your function as
g(x) = (4 - x4)^(1/2)

3. Oct 20, 2011

### CrossFit415

Alright

4. Oct 22, 2011

### CrossFit415

I have

g'(x) = 1/2(4-x)^-1/2 • -4x^3

Then how would I determine the domains? I know 4-x^4 >= 0. So what about the two square roots?

What should I do with the +/-√2 ?

Last edited: Oct 22, 2011
5. Oct 22, 2011

### Staff: Mentor

It's still not right. Check your work.

Since your original is named g, its derivative is g', not G'.

The domain of g is the interval [-$\sqrt{2}$, $\sqrt{2}$]. The domain of the this derivative will be exactly the same, with the possible exception of the endpoints.

6. Oct 22, 2011

### HallsofIvy

Staff Emeritus
NO. Do it again and show each step.

The domain will be the intersection of the natural domain of the derivative (not what you have above) and the given domain of the function, $[-\sqrt{2}, \sqrt{2}]$.

7. Oct 22, 2011

### CrossFit415

g'(x) = 1/2(4-x^4)^1/2-1 • d/dx (4-x^4)
= 1/2(4-x^4)^-1/2 • -4x^3
Should I keep continuing?

Then
= -2x^3(4-x^4)^-1/2

Last edited: Oct 22, 2011
8. Oct 22, 2011

### HallsofIvy

Staff Emeritus
Please do NOT use "x" both as the variable and for multiplication!

9. Oct 22, 2011

### CrossFit415

Sorry!

10. Oct 22, 2011

### Staff: Mentor

Much better.

When you write expressions on a single line, you need to use more parentheses. The above should be written as -2x^3(4-x^4)^(-1/2).

11. Oct 22, 2011

### CrossFit415

Thanks! Ohh ok I ll be sure to use more parenthesis next time.

So should I just plug in +/- √2 for x to find the domains? Or leave it alone? Are the domains just +/- √2 ?

12. Oct 23, 2011