# A quick simple derivatives question

1. Jun 15, 2014

### comiconor

I've never really noticed whether this is true but if I know that:

dt/ds = k/(1-s/r)

for example, how does one find ds/dt? Is it the inverse, as you would expect, or is there some other method?

2. Jun 15, 2014

### Fredrik

Staff Emeritus
The formula for the derivative of the inverse of a differentiable function f looks like this:
$$(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}.$$ So if $t=f(s)$, $s=f^{-1}(t)$ and
$$f'(s) =\frac{dt}{ds} =\frac{k}{1-\frac{s}{r}},$$ we have
$$\frac{ds}{dt} =(f^{-1})'(t) = \frac{1}{f'(f^{-1}(t))} = \frac{1}{f'(s)} = \frac{1-\frac s r}{k}.$$ So it turns out that the super-naive calculation
$$\frac{ds}{dt} =\frac{1}{\frac{dt}{ds}} = \frac{1}{\frac{k}{1-\frac{s}{r}}} =\frac{1-\frac s r}{k}$$ would have worked.

I haven't really thought about the exact circumstances in which this works. If you want to be safe, you should always rewrite things so that you can apply the formula at the start of this post.

3. Jun 15, 2014