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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 $(\sqrt{r^2 +1})^{-3/2}$

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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
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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: $$\begin{align*} \frac{d}{dr}\frac{r}{\sqrt{r^2+1}} &= \frac{\sqrt{r^2+1}-\frac{r^2}{\sqrt{r^2+1}}}{r^2+1}\\ &= \frac{\quad\frac{r^2+1}{\sqrt{r^2+1}} - \frac{r^2}{\sqrt{r^2+1}}\quad}{r^2+1}\\ &= \frac{\quad\frac{r^2+1-r^2}{\sqrt{r^2+1}}\quad}{r^2+1}\\ &= \frac{\quad\frac{1}{\sqrt{r^2+1}}\quad}{r^2+1}\\ &= \frac{1}{(r^2+1)\sqrt{r^2+1}}\\ &= \frac{1}{(r^2+1)^{3/2}}\\ &= (r^2+1)^{-3/2}. \end{align*}$$

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

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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.

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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
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