Prove using the limit definition of a derivative an equation relating total and partial derivatives of a two variable function 
*

*I've been trying to prove the following equation which relates the total derivative to the partial derivatives of a two variable function $$V(r, h)$$, namely $$\frac{dV}{dr} = \frac{\partial V}{\partial r} + \frac{\partial V}{\partial h}\frac{dh}{dr} $$ using only limits. This isn't a homework question but just something I want to learn how to do.


*I was able to do so for the special case of $$V(r, h) = \frac{\pi r^2h}{3}$$ as follows (volume of a cone with height h and radius r):
$$\frac{dV}{dr} = \lim\limits_{\Delta r \to 0} \frac{V(r+\Delta r, h(r+\Delta r))-V(r,h(r))}{\Delta r} 
\\=\lim\limits_{\Delta r \to 0} \left(\frac{\pi(r+\Delta r)^2h(r+\Delta r)}{3\Delta r} - \frac{\pi r^2}{3\Delta r}\right) 
\\=\lim\limits_{\Delta r \to 0} \frac{\pi r^2h(r+\Delta r) + 2\pi r\Delta h(r+\Delta r)+\pi(\Delta r)^2h(r+\Delta r) - \pi r^2h(r)}{3\Delta r} 
\\=\lim\limits_{\Delta r \to 0} \left(\frac{\pi r^2}{3}\frac{h(r+\Delta r)-h(r)}{\Delta r}\right) + \lim\limits_{\Delta r \to 0} \frac{2\pi rh(r+\Delta r)}{3} + \lim\limits_{\Delta r \to 0} \frac{\pi(\Delta r)h(r+\Delta r)}{3}
\\=\frac{\pi r^2}{3}\frac{dh}{dr}+\frac{2\pi rh(r)}{3} + 0
\\= \frac{\partial V}{\partial h}\frac{dh}{dr} + \frac{\partial V}{\partial r}
$$


*I wasn't able to get anywhere with an arbitrary function. The thing that's making it difficult for me is there being two variables h and r in the limit expression. I'm thinking I have to do a change of variables or use the composition limit law but it's just a guess and I haven't been making any progress.
$$\frac{dV}{dr} = \frac{\partial V}{\partial r} + \frac{\partial V}{\partial h}\frac{dh}{dr}$$ which in limit form is:
$$
\lim\limits_{\Delta r \to 0} \frac{V(r+\Delta r, h(r+\Delta r))-V(r,h(r))}{\Delta r} 
\\=\lim\limits_{\Delta r \to 0} \frac{V(r+\Delta r, h)-V(r,h)}{\Delta r} 
+\lim\limits_{\Delta h \to 0} \frac{V(r, h+\Delta h)-V(r,h)}{\Delta h}\lim\limits_{\Delta r \to 0} \frac{h(r+\Delta r)-h(r)}{\Delta r}
$$
which is what I want to prove.
Help would be much appreciated.
-Jay
 A:  derivative has a characterization that if $g : A \subseteq \mathbb{R} \to \mathbb{R}$ is differentiable at a point $x \in A$, then
$$g(x + h) = g(x) + g'(x)h + o(h),$$
where $r(h) = o(q(h))$ means $\frac{r(h)}{q(h)} \to 0$ as $h \to 0$
Now assume $f : A \subseteq \mathbb{R^2} \to \mathbb{R}$, and $x : B \subseteq \mathbb{R} \to \mathbb{R}$ and $y : B \subseteq \mathbb{R} \to \mathbb{R}$. Then
\begin{align}
f(x(t + h), y(t + h)) &= f(x + x'h + o(h), y + y'h + o(h)) \\
&= f(x + x'h + o(h), y) + f_y(x + x'h + o(h), y)(y'h + o(h)) + o(y'h + o(h)) \\
&= f(x + x'h + o(h), y) + f_y(x + x'h + o(h), y)(y'h + o(h)) + o(h) \\
&= f(x, y) + f_x(x, y)(x'h + o(h)) + o(x'h + o(h)) + f_y(x + x'h + o(h), y)(y'h + o(h)) + o(h) \\
&= f(x, y) + f_x(x, y)(x'h + o(h)) + o(h) + f_y(x + x'h + o(h), y)(y'h + o(h)) + o(h) \\
&= f(x, y) + f_x(x, y)(x'h + o(h)) + f_y(x + x'h + o(h), y)(y'h + o(h)) + o(h)
\end{align}
Now subtract $f(x, y)$ and divide by $h$ and let $h \to 0$ to get
$$\frac{d}{dt}f(x, y) = f_x(x, y)x'(t) + f_y(x, y)y'(t).$$
Whenever I omitted the argument of $x$ and $y$ I mean it to be $t$.
There is a flaw, in that I assumed $f_y$ to be continuous at $(x(t), y(t))$. I'm not sure how to avoid this or if this is necessary for the result to hold. 
The answer is wrong.
