Question

Proof of the derivative of \(f(x)= \sqrt{x}\) (see inside)

Answer 1

does this involve the limit definition of the derivative?

Answer 2

Looking to make a formal proof that the derivative of \(f(x)=\sqrt{x}\) is \(\dfrac{1}{2\sqrt{x}}\). So, to start, we plug into the formula and get \[\lim_{h\to 0} \dfrac{\sqrt{x-h}-\sqrt{x}}{h}\\
\]
Then we probably multiply by \(\dfrac{\sqrt{x}}{\sqrt{x}}\), but I get lost after that.

Answer 3

typo, that should be x+h, not x-h

Answer 4

no you have to multiply by the whole conjugate on top and bottom
\[\sqrt{x+h} +\sqrt{x}\]

Answer 5

Ohh, okay, let me play with that for a minute.

Answer 6

Okay, that's what I was missing, got it figured out.
\[
\lim_{h\to0}\dfrac{\sqrt{x+h}-\sqrt{x}}{h}\cdot\dfrac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} \\
\lim_{h\to0}\dfrac{h}{h\sqrt{x+h}+h\sqrt{x}}\\
\lim_{h\to0}\dfrac{1}{\sqrt{x+h}+\sqrt{x}}\\
\dfrac{1}{\sqrt{x}+\sqrt{x}}\\
\dfrac{1}{2\sqrt{x}}
\]
Thanks for the help.

Answer 7

yep glad to help