Chain rule and a square in the denominator

Question

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

Answer 1

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

Answer 2

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

Answer 3

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.