Why do you use the chain rule when finding the derivative of a circles's area?

Question

I know that the derivative of A=pi(r^2) is A'=2pi(r)(r'), as per the chain rule, but I don't know why A'=2pi(r) wouldn't work, as per power rule?

Answer 1

The confusion stems from sloppy notation.

Let me try to be more rigorous about it.

Let A be function defined by

A(r) = πr²

Then

dA/dr = 2πr

Now let r be a function of another variable, t:

r(t) = 2t

This would be an example of a circle growing in size as time t progresses.

Define another function

B(t) = A(r(t)) = π(r(t))² = π(2t)² = 4πt²

B(t) gives us the area of the circle at time t.

Then

dB/dt = 8πt

The chain rule tells us that the identical result could be obtained by

dB/dt = dA/dr × dr/dt = 2πr × 2 = 4πt × 2 = 8t

The confusion you pointed out has several sources:

1) The same name A is used for A(r) and B(t)

2) The same notation A' is used to mean dA/dr or dA/dt.

3) It is left unspecified whether r is a variable, or a function of t.

4) High schools teach about full derivative, and only colleges then teach about partial derivatives.

The confusion with the chain rule would disappear if high schools started with partial derivatives.

But I have the feeling that this last point may only confuse you only more.

Answer 2

This comes down to a technicality.

Both equations and “conclusions” are sort of the same, depending on the branches of dependency.

If the radius is independent of any other variable, then both equations will yield the exact same result of

A’(r) = 2πr

Additionally, chain rule is able to happen in pretty much any derivative case because something "functional" could happening on the inside of the outer function.

Now, say, for instance, the radius depended on the time, t, that passes.

That means r(t) = f(t), some function, f, of t

Sort of like r is the y-axis and t is the x-axis on the xy-plane.

dA/dt = 2π * r(t) * dr/dt

dA/dr = 2π * r(t)

Interesting to note, right? The derivative of the area with respect to (wrt) t picks up the extra derivative of r wrt t; however, the derivative of the area wrt r does not, even if its dependency lies upon t, the selectiveness of the derivative can alter the final result.

I hope this helps answer your question. Please message me in the comments if you have any questions, comments, or concerns!