# Derivatives of square roots

1. Apr 13, 2004

### the1024b

How can i find the derivative of a function like this:
f(x) = sqrt( 1 - x² )

2. Apr 13, 2004

### Hurkyl

Staff Emeritus
Do you know how to write a square root with exponents?

3. Apr 13, 2004

### the1024b

(1 - x² )^(1/2) ?

4. Apr 13, 2004

### Hurkyl

Staff Emeritus
That's right! Now, you just need to apply what you know about differentiating expressions like that.

5. Apr 13, 2004

### the1024b

si will that be:
1/2((1-x²)/2)^(-1/2)

?

6. Apr 13, 2004

### HallsofIvy

Staff Emeritus
Not quite. You have one too many "1/2"s (you don't want that "/2" inside the square root and you didn't use the chain rule.

You need to multiply by the derivative of 1-x2.

Last edited: Jul 17, 2009
7. May 13, 2009

### mathsn00b

Hi,

I have a similar problem, I need to differentiate sqrt(x^2 + y^2) in terms of x and y.
Starting this I took the simple step (x^2 + y^2)^(1/2)...

My next step is a guess and I am lost after it....(1/2)(x^2 + y^2)(-1/2)....

Any help would be much appreciated.

8. May 13, 2009

### jbunniii

If by "in terms of x and y", you mean you want to calculate the partial derivatives, then for the partial derivative with respect to x, treat y as a constant and differentiate with respect to x as you normally would a function of one variable. For the partial derivative with respect to y, treat x as constant.

9. May 13, 2009

### mathsn00b

thanks, would I do this by...

df/dx = 1/2(x^2 + y^2)^(-1/2).2x = x/sqrt(x^2 + y^2) and...

df/dy = 1/2(x^2 + y^2)^(-1/2).2y = y/sqrt(x^2 + y^2) ?

thanks for your help so quickly.

10. May 13, 2009

### jbunniii

Looks good to me.

11. Jul 16, 2009

### 68Pirate

What if i have a problem similar to these however now its 4/ ^5sqrt(x^5)

12. Jul 17, 2009

### Matthollyw00d

If that is meant to be 4^(5(sqrt(x^5))), then you can easily rewrite this to equal
4^(5(x^(5/2)) And using what you know from differentiating exponentials and chain rule, you should be able to get the rest.