Derivatives of composition function for increase rate I have composition functions $V(t)$, $r(t)$, $P(t)$, $L(t)$ and $y(t)$. And I have $\frac{d r}{d t}=0.05$, $\frac{d P}{d t}=-0.15$, $\frac{d L}{d t}=0.25$ and $\frac{d y}{d t}=-0.08$, which mean the increase rate of $r$, $P$, $L$ and $y$ by each hour $t$, and 
$$V=\frac{\pi r^3 P}{yL}$$
How can I get the increase rate of $V$ for each $t$?
It is easy for me to set $r_0$, $P_0$, $L_0$ and $y_0$ and get $V_0$ in interms of $r_0$, $P_0$, $L_0$, $y_0$ and set $r_1=1.05r_0$, $P_1=0.85P_0$, $L_1=1.25L_0$ and $y_1=0.92y_0$ and get $V_1$ to get the result. However, it is not a good way to enhence my understand of derivatives.
 A: Just apply the rules for the computation of the derivatives:
$$
\frac{dV}{dt} = \pi \frac{\left(3r^2\frac{dr}{dt}P + r^3\frac{dP}{dt}\right)yL - r^3P\left(\frac{dy}{dt}L+y\frac{dL}{dt}\right)}{y^2L^2}.
$$
To compute the derivative of $V$ at a certain time $t_\star$ you should have instantaneous information about the four functions at $t_\star$ and about their derivatives. I suppose your data are the values of the derivatives of the four functions at a certain time, but to complete the computation you also need the value of the function at the same time.
A: Given $V=\frac{\pi r^3 P}{yL}$, where $r,P,y,L$ are the functions of $t$ and $r_t=0.05$, $P_t=-0.15$, $L_t=0.25$, $y_t=-0.08$, the derivative $V_t$ is:
$$\frac{dV}{dt}=\frac{dV}{dr}\cdot \frac{dr}{dt}+\frac{dV}{dP}\cdot \frac{dP}{dt}+\frac{dV}{dy}\cdot \frac{dy}{dt}+\frac{dV}{dL}\cdot \frac{dL}{dt}=\\
\frac{3\pi r^2P}{yL}\cdot 0.05+\frac{\pi r^3}{yL}\cdot (-0.15)-\frac{\pi r^3P}{y^2L}\cdot (-0.08)-\frac{\pi r^3P}{yL^2}\cdot 0.25.$$
A: Well, if you had $V=r^2P$ then by product rule you could write
$$\frac{dV}{dt} = 2r\frac{dr}{dt}P + r^2\frac{dP}{dt}.$$
Then you can plug in whatever values you have on the right side.
