Dropping inconsequential terms From $V = \pi r^2h$, we get $\displaystyle{\frac{\partial V}{\partial r} = 2\pi rh, \ \frac{\partial V}{\partial h} = \pi r^2}$.
If $r, h$ change at the same time, we have $V + \delta V = \pi(r + \delta r)^2(h + \delta h)$ which implies $\delta V = \pi(2rh\delta r + h(\delta r)^2 + r^2\delta h + 2r\delta r \delta h + (\delta r)^2 \delta h) \approx \pi(2rh\delta r + r^2 \delta h)$ meaning $\delta V \approx \displaystyle{\frac{\partial V}{\partial r}\delta r + \frac{\partial V}{\partial h}\delta h}$.
There's a precise relation $\displaystyle{\delta z =\frac{\partial z}{\partial x}\delta x + \frac{\partial z}{\partial y}\delta y}$ for $z = f(x, y)$ whose proof is an application of chain rule.
Now suppose we do not know the latter and decide to simply compute $\delta V$ like it's done above. How do we know which terms in the expansion of $\delta V$ to drop to get the approximation? We drop the smallest ones, but how do we know, say, $r^2\delta h \not \approx (\delta r)^2\delta h?$
 A: In general we have for $f(x,y)$ sufficiently smooth
$$\begin{align}
\Delta f(x,y) &= f(x+\Delta x,y+\Delta y)-f(x,y)\\\\
&=f_1(x,y) \Delta x+f_2(x,y)\Delta y \\\\
&+f_{11}(x,y)(\Delta x)^2+f_{12}(x,y)(\Delta x)(\Delta y)+f_{22}(x,y)(\Delta y)^2+\text{Higher Order Terms}
\end{align}$$


Now, one might ask the relative sizes of the linear terms, $\Delta x$ and $\Delta y$, to the quadratic terms, $(\Delta x)^2$, $(\Delta x)(\Delta y)$, and $(\Delta y)^2$.
If $\Delta x$ and $\Delta y$ are both regarded heuristically as arbitrarily "small" numbers, then certainly $(\Delta x)(\Delta y)$ is "much smaller" than both $(\Delta x)$ and $(\Delta y)$.  And of course $(\Delta x)^2$ is "much smaller" than $\Delta x$ and $(\Delta y)^2$ is "much smaller" than $\Delta y$.  This is the reason for "dropping" the higher-order terms.


But wait!  What if $\Delta y $ is "much smaller" than $ (\Delta x)^2$.  Well if this is true, then certainly $\Delta y$ is also "much smaller" than $\Delta x$.  All this means is that even the linear term in $\Delta y$ is negligible.
And if $\Delta x $ is "much smaller" than $ (\Delta y)^2$, then certainly $\Delta x$ is also "much smaller" than $\Delta y$.  And all this means is that even the linear term in $\Delta x$ is negligible.


Since $\Delta x$ and $\Delta y$ are arbitrary, we don't assign relative sizes.  This mandates that we keep both linear terms, even though one of them might be negligible.
