find the derivative of a square root?

coolbeans33

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how do you get the derivative of g(x)= square root of 1+2x? please explain the steps for how you got the answer. my book says that it's 1/(sqrt 1+2x). this doesn't make any sense to me.
it's also asking for the domain of g(x) and the domain of g'(x). I thought the domain for g(x) would be x is greater than or equal to zero, but apparently I'm wrong.
 
how do you get the derivative of g(x)= square root of 1+2x? please explain the steps for how you got the answer. my book says that it's 1/(sqrt 1+2x). this doesn't make any sense to me.
it's also asking for the domain of g(x) and the domain of g'(x). I thought the domain for g(x) would be x is greater than or equal to zero, but apparently I'm wrong.

Can you determine f'(x) when f(x) = √x ?

For domain, investigate for what values of x, you have √(1+2x) ≤ 0 .
 
how do you get the derivative of g(x)= square root of 1+2x? please explain the steps for how you got the answer.
Please explain the steps you have tried so far.

it's also asking for the domain of g(x) and the domain of g'(x). I thought the domain for g(x) would be x is greater than or equal to zero, but apparently I'm wrong.
Try using what you learned back in algebra. When is a square root defined? In particular, what must be the sign on the insides of the square root? Then what inequality can you create with the insides of this particular square root? When you solve that, what do you get?

Please be complete. Thank you! ;)
 
how do you get the derivative of g(x)= square root of 1+2x? please explain the steps for how you got the answer. my book says that it's 1/(sqrt 1+2x). this doesn't make any sense to me.
Are you supposed to use the limit definition of the derivative?

\(\displaystyle \displaystyle g'(x) = \lim_{h \to 0} \dfrac{g(x+h) - g(x)}{h}\)

.......\(\displaystyle \displaystyle = \lim_{h \to 0} \dfrac{\sqrt{1 + 2x + 2h} - \sqrt{1 + 2x}}{h}\)

The tricky step is to rationalize the numerator by multiplying top and bottom of the fraction by the SUM of the two square roots.

Give it a try, and show us how far you get (if you are still stuck).
 
Are you supposed to use the limit definition of the derivative?

\(\displaystyle \displaystyle g'(x) = \lim_{h \to 0} \dfrac{g(x+h) - g(x)}{h}\)

.......\(\displaystyle \displaystyle = \lim_{h \to 0} \dfrac{\sqrt{1 + 2x + 2h} - \sqrt{1 + 2x}}{h}\)

The tricky step is to rationalize the numerator by multiplying top and bottom of the fraction by the SUM of the two square roots.

Give it a try, and show us how far you get (if you are still stuck).

why did you insert 2h instead of just h?
 
He didn't "insert" either h or 2h. He replaced "x" with "x+ h". That's what "f(x)" and "f(x+ h)" mean! \(\displaystyle f(x)= \sqrt{1+ 2x}\) so \(\displaystyle f(x+h)= \sqrt{1+ 2(x+h)}= \sqrt{1+ 2x+ 2h}\).
 
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