Derivatives of square roots


by the1024b
Tags: derivatives, roots, square
the1024b
the1024b is offline
#1
Apr13-04, 06:52 PM
P: 5
How can i find the derivative of a function like this:
f(x) = sqrt( 1 - x )
Phys.Org News Partner Science news on Phys.org
NASA's space station Robonaut finally getting legs
Free the seed: OSSI nurtures growing plants without patent barriers
Going nuts? Turkey looks to pistachios to heat new eco-city
Hurkyl
Hurkyl is offline
#2
Apr13-04, 06:57 PM
Emeritus
Sci Advisor
PF Gold
Hurkyl's Avatar
P: 16,101
Do you know how to write a square root with exponents?
the1024b
the1024b is offline
#3
Apr13-04, 07:02 PM
P: 5
(1 - x )^(1/2) ?

Hurkyl
Hurkyl is offline
#4
Apr13-04, 07:06 PM
Emeritus
Sci Advisor
PF Gold
Hurkyl's Avatar
P: 16,101

Derivatives of square roots


That's right! Now, you just need to apply what you know about differentiating expressions like that.
the1024b
the1024b is offline
#5
Apr13-04, 07:26 PM
P: 5
si will that be:
1/2((1-x)/2)^(-1/2)

?
HallsofIvy
HallsofIvy is offline
#6
Apr13-04, 07:56 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,885
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
mathsn00b is offline
#7
May13-09, 08:58 AM
P: 2
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
jbunniii is online now
#8
May13-09, 10:26 AM
Sci Advisor
HW Helper
PF Gold
jbunniii's Avatar
P: 2,908
Quote Quote by mathsn00b View Post
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
mathsn00b is offline
#9
May13-09, 10:40 AM
P: 2
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
jbunniii is online now
#10
May13-09, 10:52 AM
Sci Advisor
HW Helper
PF Gold
jbunniii's Avatar
P: 2,908
Quote Quote by mathsn00b View Post
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
68Pirate is offline
#11
Jul16-09, 11:50 AM
P: 2
What if i have a problem similar to these however now its 4/ ^5sqrt(x^5)
Matthollyw00d
Matthollyw00d is offline
#12
Jul17-09, 06:03 AM
P: 92
Quote Quote by 68Pirate View Post
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.


Register to reply

Related Discussions
square roots and fractions Precalculus Mathematics Homework 7
Square roots Introductory Physics Homework 7
square roots Brain Teasers 12
Complex square roots Calculus 4
Square Roots General Math 6