Using the difference quotient in the limit definition, find the derivative of (See below)

Question

anonymous · Answer

$$y=\sqrt{2x-1}$$

myininaya · Answer

$$f'(x)=\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}$$

We need to know what f(x+h) is.

We are given 

$$f(x)=\sqrt{2x-1} => f(x+h)=\sqrt{2(x+h)-1}$$

so we have


$$f'(x)=\lim_{h \rightarrow 0} \frac{\sqrt{2(x+h)-1}-\sqrt{2x-1}}{h}$$

myininaya · Answer

now rationalize the numerator by multiplying top and bottom by the conjugate of the top

myininaya · Answer

like this:
$$f'(x)=\lim_{h \rightarrow 0} \frac{\sqrt{2(x+h)-1}-\sqrt{2x-1}}{h} \cdot \frac{\sqrt{2(x+h)-1}+\sqrt{2x-1}}{\sqrt{2(x+h)-1}+\sqrt{2x-1}}$$

anonymous · Answer

Correct.

turingtest · Answer

sorry to but in but how are you with Tayor series myininaya? http://openstudy.com/#/updates/4ecd4838e4b04e045aee647b

myininaya · Answer

$$f'(x)= \lim_{h \rightarrow 0}\frac{(2(x+h)-1)-(2x-1)}{h(\sqrt{2(x+h)-1}+\sqrt{2x-1})}$$

myininaya · Answer

simplify the top and you will able to cancel something