0
$\begingroup$

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!

$\endgroup$
2
  • 1
    $\begingroup$ The derivative is a limit! $\endgroup$ Feb 5 '14 at 8:20
  • $\begingroup$ I will edit the question, Martín-Blas, to include a limit, as I'm sure that it was intended. $\endgroup$ Feb 5 '14 at 8:40
1
$\begingroup$

$$ f'(3)=\lim_{h\to 0}{\displaystyle{1\over\sqrt{3+h}}-{1\over\sqrt{3}}\over h}= \lim_{h\to 0}{\displaystyle{1\over\sqrt{3+h}}-{1\over\sqrt{3}}\over h} {\displaystyle{1\over\sqrt{3+h}}+{1\over\sqrt{3}}\over\displaystyle{1\over\sqrt{3+h}}+{1\over\sqrt{3}}}= $$

$$ \lim_{h\to 0}{\displaystyle{1\over{3+h}}-{1\over{3}}\over h}{1\over\displaystyle{1\over\sqrt{3+h}}+{1\over\sqrt{3}}}= \cdots $$

Can you continue?

$\endgroup$
2
  • $\begingroup$ I think I must be doing something wrong as I get down to: (square root x) - (square root of x + h)/ h (square root x) times by (square root x + h) Then with the top line that becomes (square root x - square root of x + h) times by (square root x + square root of x + h) This then goes to: x - x + h / h (square root x) (square root x + h) This then cancels down to 1/ (square root x) (square root x + h) If I sub x in I still don't get -1/2. Would you be able to provide a little more guidance? Thank you. :) $\endgroup$
    – Guest
    Feb 5 '14 at 8:37
  • $\begingroup$ Sorry, there really should be more formatting in that response, but I am not sure how to do that. Hope you can follow it. x $\endgroup$
    – Guest
    Feb 5 '14 at 8:46
0
$\begingroup$

Hint: to work this out in the particular case of $f(x)=1/\sqrt x$, you could start looking for the limit $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\left(f(x+h)+f(x)\right)= \lim_{h\to0}\frac{f(x+h)^2-f(x)^2}{h}.$$

$\endgroup$
1

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.