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I just studied the chain rule from Video on youtube (https://www.youtube.com/watch?v=U5_pClLuJj0)

it says: $$ frac{dy}{dx} = frac{dy}{du} * frac{du}{dx} $$

Then in the problem there is a cone and water in it. The equation for finding volume of water in the cone is $$ V = frac{pi}{3}*r^2*h $$

suppose the water draining over time, I want to find the relationship of the rate of change of Volume, radius, and height with respect to time. So There is $ V(t), r(t), h(t) $ so there rate of change with respect to time is $frac{dV}{dt}, frac{dr}{dt}, frac{dh}{dt} $

Then the master in the video says:

Since: $$ V = frac{pi}{3}*r^2*h $$

Then: $$frac{dV}{dt} = frac{d}{dt} [frac{pi}{3}*r^2*h] $$

Then: $$ frac{dV}{dt} = frac{pi}{3} [r^2 * frac{dh}{dt} + h(2r) * frac{dr}{dt}] $$

this is when I get confused because in this case r^2 and h are both function of t, so according to the product rule of derivative $frac{d}{dx} [f(x) * g(x)] = f(x)*frac{d}{dx}[g(x)] + g(x) * frac{d}{dx}[f(x)]$

in my solution it suppose to be just: $ frac{dV}{dt} = frac{pi}{3}[r^2 * frac{dh}{dt} + h(2r)]$ (with no $* frac{dr}{dt}$ at the end) or if we think of r as a function of t, why don't we also think of h as function of t? in this case we would get $ frac{dV}{dt} = frac{pi}{3}[r^2 * frac{dh}{dt} * frac{dh}{dt} + h(2r) * frac{dr}{dt}] $

I'm newbie to calculus so please explain me in detail about how the product rule and the chain rule relate to this problem in very detail. I don't see much connection right now ^^" I feel quite lost.

alright. don't worry man it is confusing at first for all but gets better with practice.what i get from question is that u think only one of (h) or (r)
are function of (t), but they are both functions of (t). u see cone is not on it's base, it's an inverted cone and water is going out from it's head like a V shape with hole in on the head. so both radius and height of water are functions of time because water is coming down to the head of cone and with decrease in height we also have decrease in radius. now we want to find the rate of decrease of volume of water when we know the rate of decrease in height and radius. in fact u must know the rate of two of three parameters {radius,height,volume} to find the other one from those two known.as u mentioned we must use chain rule $$ frac{dy}{dx} = frac{dy}{du} * frac{du}{dx} $$
along with derivative of the products:

$$frac{d}{dx} [f(x) * g(x)] = f(x)*frac{d}{dx}[g(x)] + g(x) * frac{d}{dx}[f(x)]$$
so rewriting it gives us this:
$$frac{d}{dt} [frac{pi}{3}*r^2*h] = frac{d}{dr} [frac{pi}{3}*r^2*h]timesfrac{dr}{dt} + frac{d}{dh} [frac{pi}{3}*r^2*h]timesfrac{dh}{dt}$$
no by calculating derivatives and substituting we have this:
$$frac{d}{dr} [frac{pi}{3}*r^2*h] = frac{2pi}{3}*r*h$$
$$frac{d}{dh} [frac{pi}{3}*r^2*h] = frac{2pi}{3}*r^2$$
$$frac{d}{dt} [frac{pi}{3}*r^2*h] = frac{2pi}{3}*r*htimesfrac{dr}{dt} + frac{2pi}{3}*r^2timesfrac{dh}{dt}$$
or
$$frac{dV}{dt} = frac{2pi}{3}*r*htimesfrac{dr}{dt} + frac{2pi}{3}*r^2timesfrac{dh}{dt}$$
now u must know the rate of change of (r) and (h) to find the rate of change of V. and u must also know the value of (r) and (h) at the given time.

another possibility is that u can also express radius and height in terms of time with an explicit relation for example:
$$r = 2*(5-t) ; for ; 0<t<5 $$
$$h = frac{5}{3}*(5-t)^2 ; for ; 0<t<5 $$
witch describe water ran out at time t=5. this explains rate of decrease for radius is linear while for height is quadratic and u can find the amount of volume in terms of time and also rate of decrease of volume in terms of t.
$$V = frac{pi}{3}*r^2*h = frac{pi}{3} * (2*(5-t))^2*(frac{5}{3}*(5-t)^2) = frac{20pi}{9}*(5-t)^4 ; for ; 0<t<5 $$

but because this explicit relation in terms of time is not available in most of time and u just know the rate of changes ($frac{dr}{dt} ; and ; frac{dh}{dt}$) and the values of radius and height ($r ; and ; h$) then u must use chain rule and derivative of product rule to connect them altogether.