Deferentiate the Square root of X ""using the difference quotient formula

[Chris1901]

Hi,
I am teaching myself calculus so I have no one else to ask so I need just a little help understanding one aspect of the formula for the question F(x)=sqrt(x).
I am learning to use the difference quotient. I can easily differentiate this using the power rule but is not so clear with the difference quotient.
When substituting into the formula for f(x)=Sqrt(x) I have seen and know how to do this...

1. F(x)=(sqrt(x+h)-sqrt(x)) / h


I see on the calculator on wolframalpha.com an alternative to write this which is...

2. 1 / (sqrt(h+x)+sqrt(x))

which when using limits, leads to the answer of...

3. 1/(2sqrt(x))

What I want to learn is, what is the process of going from line 1. to line 2.
How is this flipped over and an addition sign in place of the subtraction sign.
Thanks for any help.
Cheers,
Chris.

 

[HallsofIvy]

Yes, the difference quotient for \(\displaystyle \sqrt{x}\) is \(\displaystyle \frac{\sqrt{x+ h}- \sqrt{x}}{h}\). I presume you know that \(\displaystyle (a- b)(a+ b)= a^2- b^2\). To get rid of the square roots in the numerator of that fraction (rationalize the numerator), multiply both numerator and denominator by \(\displaystyle \sqrt{x+ h}+ \sqrt{x}\).

The difference quotient becomes \(\displaystyle \frac{(x+ h)- x}{h(\sqrt{x+h}+ \sqrt{x})}= \frac{h}{h(\sqrt{x+ h}+ \sqrt{h})}= \frac{1}{\sqrt{x+ h}+ \sqrt{x}}\).

Taking the limit as h goes to 0, \(\displaystyle \sqrt{x+ h}\) goes to \(\displaystyle \sqrt{x}\) so the difference quotient goes to \(\displaystyle \frac{1}{2\sqrt{x}}\). Notice that you can write that as \(\displaystyle \frac{1}{2}\frac{1}{\sqrt{x}}= \frac{1}{2}x^{-1/2}\).

 

[Chris1901]

Thanks for that great explanation.
I hadn't thought of rationalising the numerator.
I thought it would be something simple.
Cheers and thanks once again.

 

[Chris1901]

I hadn't thought about rationalising the numerator and yep noticed how it goes back into power notation.
Thanks heaps, greatly appreciated.