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.