Implicit Differentiation for calculating area of a circle

Question

I am working on a implicit differentiation equation of a circle where the radius grows by 1/per sec.

Can someone advise if I have differentiate the equation correctly?

The radius of the circle grows by 1/s given: $$\frac{dr}{dt} = 1$$

The area of the circle given: $$A = \pi r^2$$

Differentiate both sides of the equation with respect to the third variable: time $$\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$$

The differentiation of the left hand side of the equation: $$\frac{d}{dt}[A] = \frac{dA}{dt}$$

The differentiation of the right hand side of the equation: $$\frac{d}{dt} [\pi r^2] = [f(x)g'(x)+g(x)f'(x)]\frac{dr}{dt}$$

$$= [\pi\frac{d}{dr}[r^2] + r^2 \frac{d}{dr}[\pi]]\frac{dr}{dt}$$

$$= 2\pi r \frac{dr}{dt}$$

$$= 2\pi r$$ (since dr/dt = 1)

Answer 1

Since $\pi$ is a constant, all you have to do is $$ \frac {dA}{dt} = 2 \pi r\frac{dr}{dt}$$

You do not need to do the product rule in this case.