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Section 6.6 Properties of Differentials

Both derivatives and differentials (and, in fact, all forms of differentiation that you may learn about in the future) satisfy two basic properties: linearity and the product rule.


You probably use a property of ordinary derivatives, called linearity, without even thinking about it. Linearity meansthat the derivative of a sum is the sum of the derivatives (distributivity), and constants pull through the derivative. For example:

\begin{equation} \frac{d}{dx} (3x^2+e^{kx}) = \frac{d}{dx} (3x^2) + \frac{d}{dx}(e^{kx}) = 6x + k e^{kx}.\tag{6.6.1} \end{equation}

This same property applies to differentials. If \(u\) and \(v\) are any two quantities, then linearity says that

\begin{equation} d(au+bv) = a\, du + b\,dv\tag{6.6.2} \end{equation}

where \(a\) and \(b\) are constants.

In these expressions, \(u\) and \(v\) may or may not be related. For example, if \(u=x^2\) and \(v=e^{kx}\) both depend on \(x\text{,}\) then

\begin{align} d(3x^2+e^{kx}) \amp= d(3x^2) + d(e^{kx}) ,\notag\\ \amp= 6x\,dx + k e^{kx}\, dx = (6x + k e^{kx})\, dx. \tag{6.6.3} \end{align}

Product Rule.

The product rule for differentials follows the same pattern as the product rule for ordinary derivatives, namely

\begin{equation} d(uv) = u \,dv + v \,du .\tag{6.6.4} \end{equation}

As an example, consider the ideal gas law

\begin{equation} pV = nRT\tag{6.6.5} \end{equation}

relating the pressure \(p\text{,}\) volume \(V\text{,}\) and temperature \(T\) of an ideal gas, where \(R\) is a physical constant, and \(n\) is the number of moles of gas, which we will assume remains constant in this example. “Zapping” (6.6.5) with \(d\text{,}\) the product rule yields

\begin{equation} d(pV) = p\,dV + V\,dp = nR\,dT .\tag{6.6.6} \end{equation}

Chain Rule.

In Leibniz notation, the chain rule for ordinary derivatives is:

\begin{equation} \frac{dy}{dx} = \frac{dy}{du} \> \frac{du}{dx} .\tag{6.6.7} \end{equation}

From the point of view of the ratio of small changes, (6.6.7) is just a statement about the ordinary rules for manipulating fractions.

For differentials, the chain rule becomes:

\begin{equation} dy = \frac{dy}{du} \> \frac{du}{dx}\, dx .\tag{6.6.8} \end{equation}

We are assuming that \(dy\) and \(dx\) are “small enough” that we can treat the relationships between \(y\) and \(u\) and between \(u\) and \(x\) as linear, so the two rates of change in (6.6.8) just multiply.

Suppose you want to find the derivative of \(Q=\ln\sin(\theta^2)\text{.}\) How many functions are there here? Three; let's consider each in turn.

The first, “outer” function is the logarithm. Logarithm of what? Give it a name! So we set \(u=\sin(\theta^2)\text{,}\) and similarly \(v=\theta^2\text{.}\) We now have \(Q=\ln u\) and \(u=\sin v\text{,}\) so that

\begin{equation} dQ = d(\ln u) = \frac{1}{u}du = \frac{1}{u}\cos v\,dv = \frac{\cos v}{u} 2\theta\,d\theta .\tag{6.6.9} \end{equation}

Dividing by \(d\theta\) yields an expression for the derivative of \(Q\) with respect to \(\theta\text{,}\) namely

\begin{equation} \frac{dQ}{d\theta} = \frac{\cos v}{u} 2\theta = 2\theta\cot(\theta^2)\tag{6.6.10} \end{equation}

where it is customary to expand \(u\) and \(v\) in terms of \(\theta\) in the final answer.

This process is often described as “peeling an onion”, with each layer corresponding to a different function, represented here by a different variable.

Inverse Functions.

The rule for the derivative of inverse functions is

\begin{equation*} \frac{dx}{dy} = \frac{1}{dy/dx} \end{equation*}

which in differential notation becomes

\begin{equation*} dx=\frac{1}{dy/dx}\, dy . \end{equation*}

As an example, we derive the derivative formula for logarithms from that for the exponential function. If \(u=\ln{v}\text{,}\) then \(v=e^u\text{;}\) both of these equations represent the same relationship between \(u\) and \(v\text{.}\) If we don't know how to differentiate the first expression, we can differentiate the second, yielding

\begin{equation} dv = d(e^u) = e^u du = v\,du\tag{6.6.11} \end{equation}

from which it follows immediately that

\begin{equation} du = \frac{1}{v}dv\tag{6.6.12} \end{equation}

which was the desired derivative formula for the inverse of exponentation. Yes, it's that easy!

Implicit Differentiation.

Implicit differentiation can be accomplished in differential notation simply by “zapping” every term in an equation with \(d\text{,}\) that is, by taking the differential (not the derivative) of both sides of the equation, using the rules above.

For example, consider the problem of finding the slope of the tangent line to a circle at an arbitrary point. We have

\begin{equation} x^2 + y^2 = a^2\tag{6.6.13} \end{equation}

with \(a\) constant, and zapping each term with \(d\) yields

\begin{equation} 2x\,dx + 2y\,dy = 0\tag{6.6.14} \end{equation}

from which it is easy to derive

\begin{equation} \frac{dy}{dx} = -\frac{x}{y}\tag{6.6.15} \end{equation}

which can then be evaluated at the desired point.

Note the shift in emphasis when using differentials to compute derivatives in these examples:

  • We didn't solve for \(y\) (or \(x\)); no attempt was made to identify dependent and independent variables.

  • At each stage, “zapping” with \(d\) produced a new term with \(d\text{,}\) which is a differential, not a derivative.

  • The physical answer is almost never a differential, but rather a derivative, namely the ratio of two differentials.