# Single variable function derivative w.r.t. time?

A circular wire expands due to heat so that its radius increases with a speed of $$0.01 ms^{-1}$$. How rapidly does the area increase when the radius is 2 cm?

The solution goes like this:

Let x be the radius and y the area. Then:

$$y=\pi x^2$$

And then it goes like this: $${dy\over{dt}}=2\pi x {dx\over{dt}}$$ How is this possible to do? y is a single variable function and x is just the independent variable.

This confuseses me a lot.

No this is why you should have never been taught that $f(x)$ means $y = ...$. Both $y$ and $x$ are functions of $t$:
$$y(t) = \pi x^2(t)$$
Now differentiate both sides with respect to $t$--use the chain rule:
$$\frac{d}{dt}y(t) = \frac{d}{dt}\left(\pi x^2(t)\right) \\ \frac{dy}{dt} = 2\pi x \frac{dx}{dt}$$
$$y = f(x) \\ \frac{dy}{dt} = \frac{df}{dt} = \frac{df}{dx}\frac{dx}{dt}$$