Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm trying to find the derivative of $\dfrac{1}{\sqrt{x+5}}$ using $\displaystyle \lim_{h\to 0} \frac {f(x+h)-f(x)}{h}$


$$\begin{align*} \lim_{h\to 0} \frac{\dfrac{1}{\sqrt{x+h+5}}-\dfrac{1}{\sqrt{x+5}}}{h} &= \frac{\dfrac{\sqrt{x+5}-\sqrt{x+h+5}}{(\sqrt{x+h+5})(\sqrt{x+5})}}{h}\\\\ &= \frac{\dfrac{x+5-x-h-5}{(\sqrt{x+h+5})(\sqrt{x+5})}}{\dfrac{h}{\sqrt{x+5}}+\dfrac{h}{\sqrt{x+h+5}}} \end{align*}$$

I do not know if this is correct or not.. please help. I'm stuck.

share|cite|improve this question
You have a typo in the first nominator, it should not be $\frac1{\sqrt x}$. – Asaf Karagila Oct 4 '12 at 23:35
@AsafKaragila: Fixed, thanks! – Austin Broussard Oct 4 '12 at 23:36
up vote 2 down vote accepted

I would suggest, start it over..

We can introduce a variable, say $u:=x+5$. Then it's $$\begin{align*} \lim_{h\to 0} \frac1h\cdot \left( \frac1{\sqrt{u+h}} -\frac1{\sqrt{u}} \right) &= \lim_{h\to 0} \frac1h\cdot\left(\frac{\sqrt u-\sqrt{u+h}}{\sqrt{u(u+h)}} \right) = \\ &= \lim_{h\to 0} \frac1h\cdot\left(\frac{\sqrt u-\sqrt{u+h}}{\sqrt{u(u+h)}} \cdot \frac{\sqrt{u}+\sqrt{u+h}}{\sqrt{u}+\sqrt{u+h}} \right) = \\ &= \lim_{h\to 0} \frac1h\cdot\left(\frac{u-(u+h)}{\sqrt{u(u+h)}\left(\sqrt{u+h}+\sqrt u\right)}\right) = \\ &=\lim_{h\to 0}\frac{-1}{\sqrt{u(u+h)}\left(\sqrt{u+h}+\sqrt u\right)} =\\ &=\frac{-1}{u\cdot 2\sqrt u} \end{align*}$$ Finally, rewrite back $u=x+5$.

share|cite|improve this answer
It's a lot clearer with substitution and taking out $\frac{1}{h}$. Thank you. – Austin Broussard Oct 4 '12 at 23:59

Multiply the numerator and denominator by $\sqrt{x+5}\cdot\sqrt{x+h+5}$ and simplify. Equivalently, put the denominator over a common denominator and simplify the resulting four-story fraction; it’s exactly the same thing. It should also be a fairly automatic response to an expression like this, since it’s the most likely route to a simplification.

However, you need to fix an algebraic error in the numerator of the numerator: two of your signs are wrong.

share|cite|improve this answer

Do you know the derivative of logarithms? $$ f(x)=\frac{1}{\sqrt{x+5}}\\ Lf(x)=\log f(x)=-\frac{1}{2}\log(x+5)\\ \frac{f'(x)}{f(x)}=\bigg( -\frac{1}{2}\log(x+5) \bigg)'_x\\ f'(x)=f(x)\bigg( -\frac{1}{2}\log(x+5) \bigg)'_x\\ L_1=\lim_{h \to 0}\frac{-\frac{1}{2}\log(x+5+h)+\frac{1}{2}\log(x+5)}{h}=-\frac{1}{2}\lim_{h \to 0}\frac{\frac{1}{2}\log(x+5+h)-\frac{1}{2}\log(x+5)}{h} $$ Derivative of log function is available, e.g., here. Hence, $$ L_1=-\frac{1}{2(x+5)} $$ And, therefore $$ f'(x)=f(x)L_1=-\frac{1}{2(x+5)^{\frac{3}{2}}} $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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