PDA

View Full Version : Derivatives of square roots


the1024b
Apr13-04, 06:52 PM
How can i find the derivative of a function like this:
f(x) = sqrt( 1 - x² )

Hurkyl
Apr13-04, 06:57 PM
Do you know how to write a square root with exponents?

the1024b
Apr13-04, 07:02 PM
(1 - x² )^(1/2) ?

Hurkyl
Apr13-04, 07:06 PM
That's right! Now, you just need to apply what you know about differentiating expressions like that.

the1024b
Apr13-04, 07:26 PM
si will that be:
1/2((1-x²)/2)^(-1/2)

?

HallsofIvy
Apr13-04, 07:56 PM
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.

mathsn00b
May13-09, 08:58 AM
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.

jbunniii
May13-09, 10:26 AM
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.

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.

mathsn00b
May13-09, 10:40 AM
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.

jbunniii
May13-09, 10:52 AM
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.

Looks good to me.

68Pirate
Jul16-09, 11:50 AM
What if i have a problem similar to these however now its 4/ ^5sqrt(x^5)

Matthollyw00d
Jul17-09, 06:03 AM
What if i have a problem similar to these however now its 4/ ^5sqrt(x^5)
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.