How are implicit derivatives related? I am having a hard time understanding implicit differentiation and how different variables can be related. I have the following question:
The dimensions of $x$ and $y$ of an object are related to its volume $V$ by the formula $V=8x^2y$


*

*How is $\frac{dV}{dt}$ related to $\frac{dy}{dt}$ if $x$ is constant?

*How is $\frac{dV}{dt}$ related to $\frac{dx}{dt}$ if $y$ is constant?

*How is $\frac{dV}{dt}$ related to $\frac{dx}{dt}$ and $\frac{dy}{dt}$ if neither $x$ nor $y$ are constant.
I was thinking to solve this by finding the implicit derivative of x and y... am I on the right track?
 A: As far as I understand you are on the right track:
(1)
Treating $x$ as a constant and differentiating normally w.r.t. $t$
$\frac{dV}{dt} = 8x^{2}\frac{dy}{dt}$
(2)
Treating $y$ as a constant and differentiating normally w.r.t. $t$
$\frac{dV}{dt} = 16xy\frac{dx}{dt}$
(3)
Differentiating implicitly w.r.t. $t$
$\frac{dV}{dt} = 16xy\frac{dx}{dt} + 8x^{2}\frac{dy}{dt}$
A: The way i see it, you are talking about a function $V(t)$ when you have $x$ and $y$ who can vary with $t$, which, for didatic means, we will think as if it were time.


*

*If you have a function $V(x,y) = 8x^2 y$, and you know, for example, that $x$ is constant with respect to $t$ (i.e, $\frac{dx}{dt} = 0$); then you can assume that $V(x,y) = V(t) = 8x^2 y(t)$, where $y(t)$ can be any function of time. Then, in this case, we will have that $\frac{dV}{dt} = 8x^2 \frac{dy}{dt}$ (note that, since $x$ is constant with respect to $t$, the variation of $V$ with respect to $t$ is only related to the variation of $y$ with respect to $t$.

*It is a very similar case, so i'll leave it to you.

*In the case where neither $x$ nor $y$ are constant, then you'll have $V(t) = 8x(t)^2 y(t)$, so all you have to do is differentiate both functions with respect to $t$.
