finding derivative

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Jan 26, 2012
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how do i find the derivative of sqrt(s). they show the answer as 1/2*sqrt(s), but i am not sure how they got to that.
 
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how do i find the derivative of sqrt(s). they show the answer as 1/2*sqrt(s), but i am not sure how they got to that.

Don't know who "they" are but "they" are wrong.

\(\displaystyle \displaystyle y=\sqrt{s}\)

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

Using power rule:

\(\displaystyle \displaystyle y'=(1/2)s^{(1/2)-1}\)

\(\displaystyle \displaystyle y'=(1/2)s^{-1/2}\) or \(\displaystyle \displaystyle y'=\frac{1}{2s^{1/2}}\) or \(\displaystyle \displaystyle y'=\frac{1}{2\sqrt{s}}\)
 
how do i find the derivative of sqrt(s). they show the answer as 1/2*sqrt(s), but i am not sure how they got to that.

:)The power rule states that the derivative of x^n=nx^n-1.
For example, the derivative of x^2=2x
the derivative of x^3=3x^2
the derivative of x^4=4x^3
the derivative of x^1/2=1/2x^1/2 and so on.

I hope this helps.
 
:)The power rule states that the derivative of x^n=nx^n-1.
For example, the derivative of x^2=2x
the derivative of x^3=3x^2
the derivative of x^4=4x^3
the derivative of x^1/2=1/2x^1/2 and so on.

I hope this helps.

afrazer721, careless error: The derivative of x^(1/2) = (1/2)x^(-1/2)
 
Don't know who "they" are but "they" are wrong.

\(\displaystyle \displaystyle y=\sqrt{s}\)

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

Using power rule:

\(\displaystyle \displaystyle y'=(1/2)s^{(1/2)-1}\)

\(\displaystyle \displaystyle y'=(1/2)s^{-1/2}\) or \(\displaystyle \displaystyle y'=\frac{1}{2s^{1/2}}\) or \(\displaystyle \displaystyle y'=\frac{1}{2\sqrt{s}}\)

alright; thanks! it seems like my answer book always skips a few steps.. :p
 
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