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Derivatives of square roots

by the1024b
Tags: derivatives, roots, square
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the1024b
#1
Apr13-04, 06:52 PM
P: 5
How can i find the derivative of a function like this:
f(x) = sqrt( 1 - x )
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Hurkyl
#2
Apr13-04, 06:57 PM
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Do you know how to write a square root with exponents?
the1024b
#3
Apr13-04, 07:02 PM
P: 5
(1 - x )^(1/2) ?

Hurkyl
#4
Apr13-04, 07:06 PM
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Derivatives of square roots

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

?
HallsofIvy
#6
Apr13-04, 07:56 PM
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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
#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
#8
May13-09, 10:26 AM
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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
#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
#10
May13-09, 10:52 AM
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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
#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
#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.


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