I needed to differentiate Y =Square root (X): is this possible?

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Hello: a quick question that I have: is differentiating a square root possible?
I needed to differentiate Y =Square root (X)
But the only notes I have are : anx^(n-1), and using this I got dy/dx = 1/2X^(-1/2)
This answer does not seem to fit in with my question, as I have only ever differentiated things like X^3 (3x^2) and simple numbers like that.
So I was basically wondering have I missed something, or is there a separate method to differentiate roots?
 
Hello: a quick question that I have: is differentiating a square root possible?
I needed to differentiate Y =Square root (X)
But the only notes I have are : anx^(n-1), and using this I got dy/dx = 1/2X^(-1/2)
This answer does not seem to fit in with my question, as I have only ever differentiated things like X^3 (3x^2) and simple numbers like that.
So I was basically wondering have I missed something, or is there a separate method to differentiate roots?

You are correct. d/dx(sqrt(x)) = 1/[2*sqrt(x)]
 
Hello: a quick question that I have: is differentiating a square root possible?
I needed to differentiate Y =Square root (X)
But the only notes I have are : anx^(n-1), and using this I got dy/dx = 1/2X^(-1/2)
This answer does not seem to fit in with my question, as I have only ever differentiated things like X^3 (3x^2) and simple numbers like that.
So I was basically wondering have I missed something, or is there a separate method to differentiate roots?
You did it right.

\(\displaystyle y = \sqrt{x} = x^{1/2}\)

Differentiate with the power rule:

\(\displaystyle \dfrac{dy}{dx} = \frac 12 x^{-1/2} = \dfrac{1}{2\ \sqrt{x}}\)
 
You did it right.

\(\displaystyle y = \sqrt{x} = x^{1/2}\)

Differentiate with the power rule:

\(\displaystyle \dfrac{dy}{dx} = \frac 12 x^{-1/2} = \dfrac{1}{2\ \sqrt{x}}\)
Thanks! That was the answer in the book- after I saw 1/2X^(-1/2) I just thought it was wrong- didn't think of making a fraction from the negative power
 
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