# Finding the derivate of a function using first principles

I want to solve an equation from first principles. The first principles equation is: $$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$$

$$f(x) = \frac{1}{\sqrt{x}} \text{ at } x= 1$$

Basically, I need to find the derivative, but I think I am getting my working out confused as the answer is $-1/2$.

Could you please show your working out so I can understand how to solve this? Thank you! :)

Also, I am having trouble understanding when using the same first principles formula how when $f(x) = 5$ the answer is $0$.

Thank you so much for your help. It is really appreciated!

• The derivative is a limit! Feb 5 '14 at 8:20
• I will edit the question, Martín-Blas, to include a limit, as I'm sure that it was intended. Feb 5 '14 at 8:40

$$f'(3)=\lim_{h\to 0}{\displaystyle{1\over\sqrt{3+h}}-{1\over\sqrt{3}}\over h}= \lim_{h\to 0}{\displaystyle{1\over\sqrt{3+h}}-{1\over\sqrt{3}}\over h} {\displaystyle{1\over\sqrt{3+h}}+{1\over\sqrt{3}}\over\displaystyle{1\over\sqrt{3+h}}+{1\over\sqrt{3}}}=$$
$$\lim_{h\to 0}{\displaystyle{1\over{3+h}}-{1\over{3}}\over h}{1\over\displaystyle{1\over\sqrt{3+h}}+{1\over\sqrt{3}}}= \cdots$$
Hint: to work this out in the particular case of $f(x)=1/\sqrt x$, you could start looking for the limit $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\left(f(x+h)+f(x)\right)= \lim_{h\to0}\frac{f(x+h)^2-f(x)^2}{h}.$$
• And divided by $f(x+h)+f(x)$. Feb 5 '14 at 9:42