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 have to find the derivative of $r/( \sqrt{r^2 +1})$ I know that i have to start with the quotient rule so I set it up like this $(\sqrt{r^2 +1})(1) - (r) (\text{the derivative of the denominator})$ I get $1/2(r^2 +1)^{-1/2}(2r)$ or $r^3+r$ so that gives me $(\sqrt{r^2 +1}) - (r^4 + r^2)$ which does not give me the right answer of $(\sqrt{r^2 +1})^{-3/2}$

share|cite|improve this question
Alternatively, any problem involving the quotient rule can be turned into a problem involving the product rule by using negative exponents. For this example, you could rewrite the rational function as $$ r(r^2+1)^{-\frac{1}{2}}. $$ – Austin Mohr Sep 24 '11 at 20:45
up vote 2 down vote accepted

You are correct, you have to start with the quotient rule; and that the numerator of the expression you get will be $(r^2+1)(1) - r(\sqrt{r^2+1})'$; but I don't understand what you say later, and I don't see where you are dividing by the square of the numerator. So let me start from scratch.

We have: $$\begin{align*} \frac{d}{dr}\frac{r}{\sqrt{r^2+1}} &= \frac{\left(\sqrt{r^2+1}\right)(r)' - r\left(\sqrt{r^2+1}\right)'}{\left(\sqrt{r^2+1}\right)^2}\\ &= \frac{\sqrt{r^2+1} - r\left(\sqrt{r^2+1}\right)'}{r^2+1}. \end{align*}$$ (Since $(r)' = 1$, and $(\sqrt{r^2+1})^2 = r^2+1$). So we just need to figure out what the derivative of $\sqrt{r^2+1}$ is, substitute it in, and perhaps do some algebraic simplifications.

What is the derivative of $\sqrt{r^2+1}$? It's a Chain Rule, so: $$\begin{align*} \frac{d}{dr}\sqrt{r^2+1} &= \frac{d}{dr}\left(r^2+1\right)^{1/2}\\ &= \frac{1}{2}\left(r^2+1\right)^{-1/2}(r^2+1)'\\ &= \frac{1}{2}\left(r^2+1\right)^{-1/2}\left( (r^2)' + (1)'\right)\\ &= \frac{1}{2}\left(r^2+1\right)^{-1/2}\left(2r + 0\right)\\ &= \frac{2r}{2}\left(r^2+1\right)^{-1/2}\\ &= \frac{r}{\sqrt{r^2+1}}. \end{align*}$$

Now we plug that into the expression we had for the derivative of $\frac{r}{\sqrt{r^2+1}}$: $$\begin{align*} \frac{d}{dr}\frac{r}{\sqrt{r^2+1}} &= \frac{\sqrt{r^2+1} - r\left(\sqrt{r^2+1}\right)'}{r^2+1}\\ &= \frac{\sqrt{r^2+1} - r\left(\frac{r}{\sqrt{r^2+1}}\right)}{r^2+1}\\ &= \frac{\sqrt{r^2+1} - \frac{r^2}{\sqrt{r^2+1}}}{r^2+1}. \end{align*}$$ Now we do a bit of algebra. We can separate the fraction and do some simplification: $$\begin{align*} \frac{d}{dr}\frac{r}{\sqrt{r^2+1}} &= \frac{\sqrt{r^2+1}-\frac{r^2}{\sqrt{r^2+1}}}{r^2+1}\\ &= \frac{\sqrt{r^2+1}}{r^2+1} - \frac{\quad\frac{r^2}{\sqrt{r^2+1}}}{r^2+1}\\ &= \frac{(r^2+1)^{1/2}}{r^2+1} - \frac{r^2}{(r^2+1)\sqrt{r^2+1}}\\ &=\frac{(r^2+1)^{1/2}}{r^2+1} - \frac{r^2}{(r^2+1)(r^2+1)^{1/2}}\\ &= \frac{(r^2+1)^{1/2}(r^2+1)^{1/2}}{(r^2+1)(r^2+1)^{1/2}} - \frac{r^2}{(r^2+1)(r^2+1)^{1/2}} &&\text{(common denominator)}\\ &= \frac{r^2+1}{(r^2+1)^{3/2}} - \frac{r^2}{(r^2+1)^{3/2}}\\ &= \frac{r^2+1-r^2}{(r^2+1)^{3/2}}\\ &= \frac{1}{(r^2+1)^{3/2}}\\ &= (r^2+1)^{-3/2}, \end{align*}$$ or else we can do the simplification directly on the fraction we had already: $$\begin{align*} \frac{d}{dr}\frac{r}{\sqrt{r^2+1}} &= \frac{\sqrt{r^2+1}-\frac{r^2}{\sqrt{r^2+1}}}{r^2+1}\\ &= \frac{\quad\frac{r^2+1}{\sqrt{r^2+1}} - \frac{r^2}{\sqrt{r^2+1}}\quad}{r^2+1}\\ &= \frac{\quad\frac{r^2+1-r^2}{\sqrt{r^2+1}}\quad}{r^2+1}\\ &= \frac{\quad\frac{1}{\sqrt{r^2+1}}\quad}{r^2+1}\\ &= \frac{1}{(r^2+1)\sqrt{r^2+1}}\\ &= \frac{1}{(r^2+1)^{3/2}}\\ &= (r^2+1)^{-3/2}. \end{align*}$$

share|cite|improve this answer
I lost you at the part with $r^2 + 1^{1\2} on the top and then it dissapeared. – user138246 Sep 24 '11 at 20:50
@Jordan: I don't know what you are refering to. I don't have any "$1^{1/2}$. Can you tell me which display it is (first, second, third, fourth, or fifth), and which line in that display? – Arturo Magidin Sep 24 '11 at 21:04
"Now we do a bit of algebra" between the third and fourth part I do not know what you did. – user138246 Sep 24 '11 at 21:06
@Jordan: I divided: the numerator is $(r^2+1)^{1/2}$, The denominator is $r^2+1 = (r^2+1)^1 = (r^2+1)^{1/2}(r^2+1)^{1/2}$. One of the factors cancels, and you are left with just $\frac{1}{(r^2+1)^{1/2}}$. – Arturo Magidin Sep 24 '11 at 21:14
@Jordan: Again, you are being hampered because you don't have strong enough algebra skills: if $a\geq 0$, then$$a^{1/2}a^{1/2} = a^{(1/2)+(1/2)} = a^1 = a.$$ Equality is not just "one-way": if $a\gt 0$, then you can go from $a^{1/2}a^{1/2}$ to $a$, and you can go from $a$ to $a^{1/2}a^{1/2}$. – Arturo Magidin Sep 24 '11 at 21:39

You could also write the quotient as a product: $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}r}r(r^2+1)^{-1/2} &=(1)(r^2+1)^{-1/2}+r(-1/2)(r^2+1)^{-3/2}(2r)\\ &=(r^2+1)^{-1/2}-\frac{r^2}{r^2+1}(r^2+1)^{-1/2}\\ &=\frac{1}{r^2+1}(r^2+1)^{-1/2}\\ &=(r^2+1)^{-3/2} \end{align} $$

share|cite|improve this answer

I don't think you've quite applied the chain rule properly. You're trying to differentiate $\sqrt{r^2 + 1}$, which is $(r^2 + 1)^{1/2}$. The derivative of this is $\frac{1}{2}(r^2 + 1)^{\underline{\underline{-1/2}}}\times 2r$ - the double underlined bit is the bit I think you forgot.

share|cite|improve this answer
I had that but didn't post it I guess, I will fix it. – user138246 Sep 24 '11 at 20:45
@Jordan: you definitely didn't have that. All of your later working followed as if it was just $(r^2+1)$. In other news, don't forget that you have to divide by the square of the denominator to apply the quotient rule, which (on second reading) it doesn't look like you've done. – Billy Sep 24 '11 at 20:49
That is what I forgot I think. – user138246 Sep 24 '11 at 20:52

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.