Differentials in Multivariable Calculus

Question

Does the idea of composing/decomposing the fraction notation of the derivative from/into differentials apply in multivariable calculus? I realize that this practice is considered non-standard and many don't like it even in single variable calculus, but we can multiply both sides of $dy/dx = f'(x)$ by $dx$ yielding $dy = f'(x)dx$, and invert the process by dividing both sides by $dx$ to return to the original equation.

In the multivariable world, differentials are more diverse. Instead of having only one dimension in which to nudge the input value of a function, we can take an indefinitely small step in an infinite number of directions, a prime candidate being ∂x, a tiny nudge in the direction of the x-axis, as shown.

Similarly, $∂y$, $∂z$, etc., represent nudges parallel to the relevant input axes. However, the formula one comes across for the differential of $f$ in multivariable calculus is interesting.

Firstly, "full differentials" ($df$, $dy$, etc.) are present, whereas one might expect to find "partial differentials" ($∂f$, $∂y$, etc.). What is the meaning of full differentials in multivariable calculus? Secondly, dividing through by one of the differentials would yield an equation whose truth isn't obvious. For example, dividing by $dy$ results in $df/dy = f_x dx/dy + f_y + f_z dz/dy$. Again, it's not clear what $df/dy$ means when $f$ has a three variable input. Does this invite thinking about the inputs as living on three non-continuous number lines rather than in a single three-dimensional space, whereas $∂f/∂y$ would indicate the latter? What about $dx/dy$ and $dz/dy$, which are input-to-input nudge ratios; are these concerning since they don't involve the output of the function at all (i.e., it might not be possible for $z$ to be a function of $x$)? Is this equation and the others like it valid, and are they useful?

Answer 1

I think looking up "Total Differentiation" in Google will guide you! As far as I know, the equation df = fxdx+fydy+fzdz is valid and could be roughly interpreted as:

1. the change in the function f (= df) equals to the product of following:
2. fx times dx, which means change in the value of x (= dx) times change in function f "in the direction of x" (= fx)
3. fy times dy
4. fz times dz

the interpretation of number three and number four is similar to number 2.

This interpretation makes intuitive sense! For example let's suppose your weight depends on two variables, the amount of ice cream and the amount of hamburger you eat. Then the change in your weight would equal to the change in the amount of ice cream multiplicated by the change in your weight that one unit ice cream caused, plus the change in the amount of hamburger times the change in your weight that one unit of hamburger caused.

Hope this helped!

(And again I do not know this subject very well, so if you have found any errors please correct me!)

Answer 2

Note: This answer is mainly based upon Introduction to Calculus and Analysis I by R. Courant and F. John.

We start with the single-variable case and consider a function $y=f(x)$. We will see that it is also quite standard to treat $dx$ and $dy$ as separate quantities provided we use the appropriate settings.

• Definition of derivative

  The definition of the derivative appears in several different forms, using the notation of Lagrange $y^{\prime}=f^{\prime}(x)$ we write \begin{align*} f^{\prime}(x)=\lim_{x_1\to x}\frac{f(x_1)-f(x)}{x_1-x}=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \end{align*} Using the notation of Leibniz we write \begin{align*} \frac{dy}{dx}=\frac{df(x)}{dx}=f^{\prime}(x)=\lim_{x_1\to x}\frac{f(x_1)-f(x)}{x_1-x}=\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x} \end{align*}

  In Leibniz's notation the passage to the limit in the process of differentiation is symbolically expressed by replacing the symbol $\Delta$ by the symbol $d$, motivating Leibniz's symbol for the derivative defined by the equation \begin{align*} \color{blue}{\frac{dy}{dx}=\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x}}\tag{1} \end{align*}

Here we have the differences $\Delta x$ and $\Delta y$ which are separate symbols. But in order to obtain the derivative $\frac{dy}{dx}$ we have to assure that $\Delta x$ is not zero and we perform the passage to the limit by means of a transformation which also in the limit avoids division by zero. In this context and with this definition (1) we have to treat $\frac{dy}{dx}$ as single symbol which cannot be separated into two different quantities $dy$ and $dx$. But that's not the end of the story.

• Definition of differentials

The derivative of a function $y=f(x)$ was defined by \begin{align*} f^{\prime}(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x} \end{align*} where $\Delta x=h$. If for a fixed $x$ and a variable $h$, we define a quantity $\varepsilon$ by \begin{align*} \varepsilon(h)=\frac{f(x+h)-f(x)}{h}-f^{\prime}(x)=\frac{\Delta y}{\Delta x}-f^{\prime}(x), \end{align*} then the fact that $f^{\prime}(x)$ is the derivative of $f$ at the point $x$ amounts to the equation \begin{align*} \lim_{h\to 0}\varepsilon(h)=0 \end{align*} The quantity $\Delta y=f(x+h)-f(x)$ represents the change or increment in the value of the dependent variable $y$ that results when the value $x$ of the independent variable is changed by the amount $\Delta x=h$. Since \begin{align*} \Delta y=f^{\prime}(x)\Delta x+\varepsilon \Delta x, \end{align*} the quantity $\Delta y$ appears as the sum of two parts, namely, a part $f^{\prime}(x)\Delta x$ which is proportional to $\Delta x$ and a part $\varepsilon \Delta x$ which can be made as small as we please compared to $\Delta x$ by making $\Delta x$ itself small enough. The dominant, linear part in the expression for $\Delta y$ we shall call the differential $dy$ of $y$ and write for it \begin{align*} dy=df(x)=f^{\prime}(x)\Delta x\tag{2} \end{align*}

Here in (2) we can see how $dy$ becomes a symbol by its own. It is by definition and we will shortly see that this kind of definition is in harmony with Leibniz's symbol $\frac{dy}{dx}$.

• For any differentiable function $f$ and for a fixed $x$ this differential (2) is a well-defined linear function of $h=\Delta x$.

For example, for the function $y=x^2$ we have $dy=d(x^2)=2x\Delta x=2xh$.

For the particular function $y=x$ whose derivative has the constant value one, we simply have $dx=\Delta x$. It is then consistent with our definition to write $dx$ for $\Delta x$ when $x$ is the independent variable; hence the differential of any function $y=f(x)$ can also be written as \begin{align*} \color{blue}{dy=f^{\prime}(x)dx}. \end{align*}

• Summary (verbatim from R. Courant): Earlier we used the symbol $dy/dx$ purely symbolically to denote the limit of the quotient $\Delta y/\Delta x$ for $\Delta x$ tending to zero. With our present definition of the differentials $dy$ and $dx$ the derivative $dy/dx$ can actually be considered as the ordinary quotient of $dy$ and $dx$. Here, however, $dy$ and $dx$ are now not in any sense "infinitely small" quantities or "infinitesimals" such an interpretation would be devoid of meaning.

Instead $dy$ and $dx$ are well-defined linear functions of $h=\Delta x$ which for large $\Delta x$ may have large numerical values. There is nothing remarkable in the fact that the quotient $dy/dx$ of those quantities has the same value as the derivative $f^{\prime}(x)$. This is merely a tautology restating the definition of $dy$ as $f^{\prime}(x)dx$.

We will now have a short look at the multi-variable case. We consider for convenience only a bivariate function $u=f(x,y)$. As for functions of one variable we consider \begin{align*} \Delta u=f(x+h,h+k)-f(x,y)=h f_x(x,y)+kf_y(x,y)+\varepsilon_1 h+\varepsilon_2 k \end{align*}

We call the linear part the differential of the function, and write \begin{align*} du=df(x,y)=\frac{\partial f}{\partial x}h+\frac{\partial f}{\partial y}k=\frac{\partial f}{\partial x}\Delta x+\frac{\partial f}{\partial y}\Delta y\tag{3} \end{align*}

This differential, sometimes called the total differential is a function of four independent variables, namely, the coordinates $x$ and $y$ of the point under consideration and the increments $h$ and $k$ of the independent variables. It simply means, that $du$ approximates to the increment $\Delta u=f(x+h,y+k)-f(x,y)$ of the function, with an error that is an arbitrary small fraction $\varepsilon_1$ of $h$ and $\varepsilon_2$ of $k$, provided that $h$ and $k$ are sufficiently small quantities.

For the independent variables $x$ and $y$ we find from (3) that \begin{align*} dx&=\frac{\partial x}{\partial x}\Delta x+\frac{\partial x}{\partial y}\Delta y=\Delta x\\ dy&=\frac{\partial y}{\partial x}\Delta x+\frac{\partial y}{\partial x}\Delta y=\Delta y\\ \end{align*} Hence, the differential $df(x,y)$ is written more commonly \begin{align*} df(x,y)=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy=f_x(x,y)dx+f_y(x,y)dy \end{align*}

Finally we consider the total differential \begin{align*} df=f_xdx+f_ydy \end{align*} and the related expression \begin{align*} \frac{df}{dy}=f_x\frac{dx}{dy}+f_y\tag{4} \end{align*}

In (4) we have a function $f=f(x,y)$ and consider $x=x(y)$ as function of $y$, so that $f=f(x(y),y)$ is a function in $y$ and $\frac{dx}{dy}=\frac{d}{dy}x(y)$.