Differentiating functions of two variables

Question

If I have a function $u = u(x,y)$ where $y = Ce^x$, my textbook says the following:

$\frac{d}{dx}u(x, Ce^x) = \frac{\partial u}{\partial x} + Ce^x \frac{\partial u}{ \partial y}$

However I do not understand why would that be the case. Is this a consequence of the chain rule, if so then how is the expression derived?

Answer 1

This is the result of horrendous notation for the partial derivative.

A better notation is to subscript the partial differential with the variable that is being allowed to vary. Using this notation, you have, for $u = f(x, y)$,

$$ du = \partial_x u + \partial_y u $$

In other words, the changes in $u$ can be split up into the changes in $u$ that are due directly to $x$ and the changes in $u$ that are due to $y$.

We can divide both sides of the equation by $dx$, since that is the independent variable. This gives:

$$ \frac{du}{dx} = \frac{\partial_x u}{dx} + \frac{\partial_y u}{dx} $$

We can also multiply anything here by $\frac{dy}{dy}$ since that is equal to 1. This gives:

$$ \frac{du}{dx} = \frac{\partial_x u}{dx} + \frac{\partial_y u}{dx}\frac{dy}{dy} $$

Now, because they are fractions, we can rearrange the rightmost fraction as:

$$ \frac{du}{dx} = \frac{\partial_x u}{dx} + \frac{\partial_y u}{dy}\frac{dy}{dx} $$

Since $y = Ce^x$, then $\frac{dy}{dx} = Ce^x$. Therefore, we can substitute that in above, yielding:

$$ \frac{du}{dx} = \frac{\partial_x u}{dx} + \frac{\partial_y u}{dy} Ce^x $$

That is the answer your textbook gives, but in a slightly different notation.

In short, the standard notation for partial differentiation is an awful mess which confuses students to no end. Once you put a slightly different notation on it, all of the results become extremely intuitive and obvious.

I'm currently working on a paper that will help sort this out, because it is really long overdue.

Answer 2

The right-hand side is just $\dfrac{\partial u}{\partial x} + \dfrac{\partial y}{\partial x} \dfrac{\partial u}{\partial y}$, so yes, it is basically the chain rule. The logic is that $du=\dfrac{\partial u}{\partial x} dx+\dfrac{\partial u}{\partial y}dy$ and $dy=\dfrac{\partial y}{\partial x}dx$ and $du=\dfrac{d u}{d x}dx$.

Answer 3

Yes it is exactly the chain rule since

• $u = u(x,y(x))$

we have

$$\frac{d}{dx}u(x, y(x)) = \frac{\partial u}{\partial x} \frac{\partial x}{\partial x} + \frac{\partial u}{ \partial y}\frac{\partial y}{\partial x}= \frac{\partial u}{\partial x} + \frac{\partial u}{ \partial y}y’(x)$$

Answer 4

Given u = u(x(t), y(t))

$\frac{\delta u}{\delta x} := \lim_{\Delta x \to 0}\frac{u(x+\Delta x,y)-u(x,y)}{\Delta x} = \lim_{\Delta x \to 0} \frac{\Delta u}{\Delta x}$

Thus, in the limit, if y is held constant, $\Delta u =\frac{\delta u}{\delta x} \Delta x$

Similarly, if x is held constant, $\Delta u = \frac{\delta u}{\delta y}\Delta y$

For well-behaved functions, these can be combined into $\Delta u = \frac{\delta u}{\delta x}\Delta x+\frac{\delta u}{\delta y}\Delta y$, although a truly rigorous derivation gets rather complicated (taking limits within limits). At least intuitively, this should make sense: if you imagine a Cartesian plane with different elevations as u, then in the limit, the change in u is equal to the change in x times the slope in the x direction, plus the change in y times the slope in the y direction.

In vector terms, this can be thought of as $\Delta u(\textbf v) = \nabla u \cdot \Delta \textbf v$, where $\nabla u$ consists on the partial derivatives, and $\Delta \textbf v$ is the change in the input vector $ \textbf v$.

For example, if average college GPA increases 0.3 points for every extra point of high school GPA, and 0.2 points for every 100 points in SAT, then we should expect an increase of 0.1 in high school GPA and 50 points in SAT to correspond, roughly speaking, with an increase of 0.3*0.1+0.2*0.5 = 0.13 in college GPA. This can be thought of as a dot product of [0.3,0.2] and [0.1,0.5].

Just as the change in u is change in its inputs times the derivative, the change in its inputs is equal the change in their inputs times the derivative:

$\Delta x = \frac{dx(t)}{dt}\Delta t $ and $\Delta y = \frac{dy(t)}{dt}\Delta t $, so

$\Delta u = \frac{\delta u}{\delta x} \frac{dx(t)}{dt}\Delta t+\frac{\delta u}{\delta y} \frac{dy(t)}{dt}\Delta t$

Taking the limit,

$\frac{du}{dt} = \frac{\delta u}{\delta x}\frac{dx(t)}{dt}+ \frac{\delta u}{\delta y}\frac{dy(t)}{dt}$