# Thread: Differentiation from First Principals

1. ## Differentiation from First Principals

ok i dont know how to write square root so '\' means 'square root'

Q1. Differentiate by using the Chain Rule

f(x)=\2x-1

Square root of 2x-1

Q2. Differentiate by First Principals

f(x)=\2x-1

I already know that the answers should be the same but i cant work it out using first principals and keep ending up with f'(x)=1

Spoiler:

$\displaystyle \frac{\sqrt{2(x+h)-1)}-\sqrt{2x-1}}{h}$

$\displaystyle \frac{1}{h}(\sqrt{2(x+h)-1)}-\sqrt{2x-1})$

$\displaystyle \frac{1}{h}(\sqrt{2(x+h)-1)}-\sqrt{2x-1})\times \frac{\sqrt{2(x+h)-1)}+\sqrt{2x-1}}{\sqrt{2(x+h)-1)}+\sqrt{2x-1}}$

$\displaystyle \frac{1}{h}(\sqrt{2(x+h)-1)}-\sqrt{2x-1})\times \frac{\sqrt{2(x+h)-1)}+\sqrt{2x-1}}{\sqrt{2(x+h)-1)}+\sqrt{2x-1}}$

$\displaystyle \frac{1}{h}\left(\frac{2x+2h-1-2x+1}{\sqrt{2(x+h)-1)}+\sqrt{2x-1}}\right)$

$\displaystyle \frac{1}{h}\left(\frac{2h}{\sqrt{2(x+h)-1)}+\sqrt{2x-1}}\right)$

$\displaystyle \frac{2}{\sqrt{2(x+h)-1)}+\sqrt{2x-1}}$

Now

$\displaystyle h\to 0$