Using the chain rule to solve a derivative

Question

I have a known derivative given as:

$\frac{dr}{dt} = \frac{a}{r^2}$

And then from that information, I am trying to find:

$\frac{d(r^2)}{dt}$

I know that this is equal to $\frac{dr^2}{dr} \times \frac{dr}{dt}$

Which gives by the chain rule:

$\frac{2a}{r}$

But its really not obvious to me why the chain rule is done here in order to solve it. Hope some one can explain it better than what I have so far read online, it is very confusing.

I take the chain rule for a function within a function, but i am not seeing how this is the case for $r^2$.

Answer 1

Let $f$ be a differentiable function and put $g(x):=f(x)^2.$ Then $g(x)=h(f(x)),$ with $h(x)=x^2.$

The chain rule gives:

$$g'(x)=h'(f(x)) f'(x)=2f(x)f'(x).$$

Answer 2

Go back to the definition. For a small $h > 0$ $$ \frac{r^2(t+h) - r^2(t)}{h} = \frac{r^2(t+h) - r^2(t)}{r(t+h) - r(t)} \frac{r(t+h) - r(t)}{h}$$ but the first denominator is then equal to $(r(t+h) - r(t))(r(t+h) + r(t))$ hence $$ \frac{r^2(t+h) - r^2(t)}{h} = \Big( r(t+h) + r(t) \Big) \frac{r(t+h) - r(t)}{h}.$$ Taking the limit as $h \to 0$ you see right away that the first factor goes to $2r(t)$ while the second one goes to $\frac{a}{r(t)^2}$ according to your hypothesis. Hence the result.

Answer 3

The chain rule may be more intuitively implemented by simply writing $$ d(r^2) = 2 r dr = 2 r \frac{a}{r^2} dt $$ from which you deduce the derivative $$ \frac{d r^2}{dt}= \frac{2a}{r} $$