I am trying to understand a step in the theory section of my differential equations textbook. The author writes,

For example, suppose we transform the first order differential equation $\frac{dy}{dx} = f(x,y)$ by the substition $y = g(x,u)$, where $u$ is regarded as a function of the variable $x$. If $g$ possesses first-partial derivatives, then the Chain Rule gives $$\frac{dy}{dx}= \frac{\partial g}{\partial x}\frac{dx}{dx} + \frac{\partial g}{\partial u}\frac{du}{dx}, $$ or $$ \frac{dy}{dx}= g_x(x,u) + g_u(x,u)\frac{du}{dx}. $$

The author then goes on to describe a technique in differential equations. But I am having trouble understanding the application of the chain rule to this function.

I took Calculus III, and for that class when we had functions of multiple variables, we created a function $D\mathbf{g}(x,u)$ and used the chain rule with $D\mathbf{g}(x,u) = D\mathbf{g}(x,u) \cdot \mathbf{u}(x)$. Constructing matrices of partial derivatives and performing the dot product then gave the desired result.

  • Could that same process be applied to this situation?
  • If not, is there an easier way to think about taking the derivative of a function of relatively few variables, so that I don't have to make the matrices and compute the dot product each time?
  • $\begingroup$ There is an easy way to remember the chain rule for multivariable functions by visualizing a simple diagram. It's explained briefly here. $\endgroup$ – wgrenard Feb 4 '16 at 2:37
  • $\begingroup$ I've been looking through your link and it seems very helpful so far. In the proof, $\epsilon_1$ and $\epsilon_2$ appear, but I don't understand what they represent. Could you shed any light? $\endgroup$ – EternusVia Feb 4 '16 at 2:52
  • $\begingroup$ I've never seen the proof done that way before, but it seems like those terms are added because without them the formula they have written is technically incorrect. This is because the values of the derivatives of $z$ actually vary over the lengths $\Delta x$ and $\Delta y$. And that formula assumes a constant slope over the lengths of these two delta terms. To remedy this they added those two epsilon terms to represent the error that arises from assuming that the derivatives are constant. This error should vanish though as the delta terms go to zero, which is what happens in the proof. $\endgroup$ – wgrenard Feb 4 '16 at 3:16

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