A little advice on using the chain rule for differentiation

Question

I must be a little rusty, but how would I evaluate the following:

$$ \frac{d}{dr}\left(1-\frac{b(r)}{r}\right)^{-1} $$

My stickler is that $b$ is a function of $r$...

Answer 1

Apply power rule outside, then chain rule for ecpression inside bracket.

Then apply quotient rule on $\dfrac{b(r)}{r}$, then again chain rule on $b(r)$ $$\frac{d}{dr}\left(1-\frac{b(r)}{r}\right)^{-1} = -1\left(1-\frac{b(r)}{r}\right)^{-2}\cdot(-1)\cdot\left(\dfrac{r\cdot\dfrac{d}{dr}b-b(r)}{r^2}\right)$$

$$=\dfrac{r\cdot\dfrac{db}{dr}-b(r)}{(r-b(r))^2}$$

Answer 2

You could write

$$\left(1-\frac{b(r)}{r}\right)^{-1}=\left(\frac{r-b(r)}{r}\right)^{-1}=\frac{r}{r-b(r)}$$

Thus, we differentiate using the quotient rule and chain rule. We express the derivative of $b(r)$ as $b'(r)$. An alternate notation is $b'(r)=\frac{db}{dr}$.

$$\left(\frac{r}{r-b(r)}\right)'=\frac{r-b(r)-r(1-b'(r))}{(r-b(r))^2}=\frac{rb'(r)-b(r))}{(r-b(r))^2}$$