3.6: Differentials

An ideal gas satisfies the equation \(PV \;=\; RT\), where \(R\) is a constant and \(P\), \(V\), and \(T\) are the pressure, volume per mole, and temperature, respectively, of the gas. It will be proved that

\[\label{eqn:gaslaw} \dfrac{\dP}{P} ~+~ \dfrac{\dV}{V} ~=~ \dfrac{\dT}{T} ~. \]

Recall that \(\dP\), \(\dV\), and \(\dT\) represent infinitesimal changes in the quantities \(P\), \(V\), and \(T\), respectively. Notice that none of the quotients in Equation ([eqn:gaslaw]) have an infinitesimal in the denominator. For example, \(\dP\) is divided not by \(\dx\) or \(\dt\), as it would be in a derivative such as \(\frac{\dP}{\dx}\) or \(\frac{\dP}{\dt}\). Instead it is divided by \(P\), which is not an infinitesimal. So Equation ([eqn:gaslaw]) is an equation that relates infinitesimals themselves, i.e. infinitesimal changes, not infinitesimal rates of change. This is, in fact, how many physical laws are stated, for reasons that will be discussed shortly.

Though infinitesimals have been used throughout this text, many calculus textbooks⁷ do not even mention them, instead preferring to call them differentials.⁸ For compatibility, the definition is given here:

Note that this is identical to Equation ([eqn:dffprimedx]) in Section 1.3.

Example \(\PageIndex{1}\): diff1

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Solution

Find the differential \(\df\) of \(f(x) = x^3\).

Solution: By definition,

\[\df ~=~ f'(x)\;\dx ~=~ 3x^2\;\dx \nonumber \]

Equivalently, this can be written as

\[d(x^3) ~=~ 3x^2\;\dx ~, \nonumber \]

which is often the way it would appear in textbooks in the sciences.

All the rules for derivatives (e.g. sum rule, product rule) apply to differentials, and can be proved simply by multiplying the corresponding derivative rule by \(\dx\) on both sides of the equation: For example, to prove (e), multiply both sides of the usual Product Rule by \(\dx\) so that

\[\begin{aligned} \frac{d(fg)}{\dx} ~=~ f\,\frac{\dg}{\dx} ~+~ g\,\frac{\df}{\dx} \quad&\Rightarrow\quad d(fg) ~=~ \cancel{\dx}\,\left(f\,\frac{\dg}{\cancel{\dx}} ~+~ g\,\frac{\df}{\cancel{\dx}}\right)\\ &\Rightarrow\quad d(fg) ~=~ f\,\dg ~+~ g\,\df \quad\checkmark\end{aligned} \nonumber \]

since the \(\dx\) terms all cancel. The proofs of the other rules are similar.

The differential version of the ideal gas law in Equation ([eqn:gaslaw])

\[\dfrac{\dP}{P} ~+~ \dfrac{\dV}{V} ~=~ \dfrac{\dT}{T} \nonumber \]

can now be proved by taking the differential of both sides of the equation \(PV = RT\):

\[\begin{aligned} d(PV) ~&=~ d(RT) ~=~ R \cdot \dT \quad\text{by the Constant Multiple Rule}\\ V \; \dP ~+~ P \; \dV ~&=~ \frac{PV}{T} \; \dT \quad\text{by the Product Rule and since $R = \frac{PV}{T}$}\

\[6pt] \frac{V\;\dP}{PV} ~+~ \frac{P\;\dV}{PV} ~&=~ \frac{\dT}{T} \quad\text{after dividing both sides by $PV$}\

\[6pt] \frac{\dP}{P} ~+~ \frac{\dV}{V} ~&=~ \frac{\dT}{T} \quad\checkmark\end{aligned} \nonumber \]

Notice that \(\frac{\dP}{P}\), \(\frac{\dV}{V}\) and \(\frac{\dT}{T}\) represent the relative infinitesimal changes in \(P\), \(V\), and \(T\), respectively. The differential formulation is useful for finding one relative infinitesimal change when the other two are known.

Example \(\PageIndex{1}\): diff2

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Solution

Suppose that \(M\) is the total mass of a rocket and its unburnt fuel at any time \(t\) (so \(M\) is a function of \(t\)). Over an infinitesimal time \(\dt\) a mass \(\dm\) of fuel is burnt and the gas byproducts are expelled out the rear of the rocket at a velocity \(v_E\) relative to the rocket. Using the law of conservation of momentum over the interval \(\dt\), show that

\[v_E\,\dm ~=~ M\,\dv \nonumber \]

where \(m\) and \(v\) are the mass of burnt fuel and the velocity of the rocket, respectively, at the beginning of the time \(\dt\).

Solution: Momentum is defined as mass times velocity. The momentum of the rocket at the beginning of the time \(\dt\) is thus \(Mv\). At the end of the time \(\dt\), the momentum of the rocket consists of two parts, namely the momentum of the rocket and its remaining unburnt fuel, which is

\[\begin{gathered} \text{((mass before $\dt$) $-$ (increase in burnt fuel)) $\times$ ((velocity before $\dt$) $+$ (increase in velocity))}\\ (M - \dm)(v + \dv)\end{gathered} \nonumber \]

and the momentum of the fuel that was burnt and expelled out the rear, which is

\[(v - v_E)\,\dm ~. \nonumber \]

So by conservation of momentum,

\[\begin{aligned} Mv ~&=~ (M - \dm)(v + \dv) ~+~ (v - v_E)\,\dm\\ Mv ~&=~ Mv ~-~ v\,\dm ~+~ M\,\dv ~-~ (\dm)(\dv) ~+~ v\,\dm ~-~ v_E\,\dm, \quad\text{so}\\ v_E\,\dm ~&=~ M\,\dv ~-~ (\dm)(\dv) ~=~ M\,\dv\end{aligned} \nonumber \]

since \((\dm)(\dv) = (m'(t)\,\dt)(v'(t)\,\dt) ~=~ m'(t)v'(t)(\dt)^2 ~=~ m'(t)v'(t) \cdot 0 = 0\).

Dividing both sides of \(v_E\,\dm = M\,\dv\) by \(\dt\) yields the equation

\[M\,\dot{v} ~=~ \dot{m}\,v_E \nonumber \]

using the dot notation—mentioned in Section 1.3—for the derivative with respect to the time variable \(t\), which is still popular with physicists. Since \(\dot{v}\) is just acceleration \(a\), this formulation is the classic equation for the acceleration of a rocket.⁹

Letting \(f\) be the natural logarithm function and letting \(g = u\) in the differential version of the Chain Rule yields the following useful result:

This is often used in a differential version of the technique of logarithmic differentiation discussed in Section 2.3.

Example \(\PageIndex{1}\): diff3

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Solution

Prove the relation \(\frac{\dP}{P} + \frac{\dV}{V} = \frac{\dT}{T}\) using logarithmic differentiation.

Solution: Take the natural logarithm and then the differential of both sides of the equation \(PV = RT\):

\[\begin{aligned} \ln\,(PV) ~=~ \ln\,(RT) \quad&\Rightarrow\quad \ln\,P ~+~ \ln\,V ~=~ \ln\,R ~+~ \ln\,T\\ &\Rightarrow\quad d(\ln\,P ~+~ \ln\,V) ~=~ d(\ln\,R ~+~ \ln\,T)\\ &\Rightarrow\quad \frac{\dP}{P} ~+~ \frac{\dV}{V} ~=~ 0 ~+~ \frac{\dT}{T} ~=~ \frac{\dT}{T} \quad\text{(since $\ln\,R$ is a constant)}\end{aligned} \nonumber \]

Example \(\PageIndex{1}\): diff4

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Solution

The derivative of the area \(\pi r^2\) of a circle of radius \(r\), as a function of \(r\), equals its circumference \(2\pi r\). Use the notion of a differential as an infinitesimal change to explain why this makes sense geometrically.

Solution: Let \(A = \pi r^2\) be the area of a circle of varying radius \(r\). Then \(A'(r) = 2\pi r\), which is equivalent to saying \(d\negmedspace A = 2\pi r\,\dr\). To see why this makes sense geometrically, imagine increasing the radius by \(\dr\), as in the picture below on the left. This increases the area \(A\) of the circle to \(A + d\negmedspace A\), with \(d\negmedspace A\) the infinitesimal area of the shaded ring in the picture.

Slice that ring along the dashed line then roll it flat, yielding a trapezoid with height \(\dr\), top length \(2\pi r\) (from the circumference of the inner circle of the ring), and bottom length \(2\pi(r+\dr)\) (from the circumference of the outer circle of the ring), as shown in the picture above on the right. The triangular edges of the trapezoid contribute nothing to the area of the trapezoid, since (by the Microstraightness Property) the hypotenuse of each is indeed a straight line, so each is a right triangle with height \(\dr\) and (by symmetry) base \(\pi\dr\), thus having area \(\frac{1}{2}\pi(\dr)^2 = 0\). Hence the entire area \(d\negmedspace A\) of the trapezoid comes from the rectangular portion of height \(\dr\) and base \(2\pi r\), which means \(d\negmedspace A = 2\pi r\,\dr\), as expected.

The above example answers the question of whether it is a happy coincidence that the derivative of a circle’s area turns out to be the circle’s circumference—no, it is not! Some other such cases (e.g. the derivative of a sphere’s volume is its surface area) are left to the exercises. Note that a similar “coincidence” does not occur for a square: if \(x\) is the length of each side then the area is \(x^2\), but the derivative of \(x^2\) is \(2x\), which is not the perimeter of the square (i.e. \(4x\)). Why does this not follow the same pattern as the circle? Think about a key difference in the shape of a square in comparison to a circle, keeping differentiability in mind.

There are many benefits to using differentials—i.e. infinitesimals—in calculus.¹⁰ For example, recall Example

Example \(\PageIndex{1}\): relrate3

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Solution

in Section 3.5 on related rates, where the volume \(V\) of a right circular cylinder with radius \(r\) and height \(h\) changes with time \(t\) as

\[\dVdt ~=~ \left( 2\pi \,r\;\cdot\;\drdt\right)\,h ~+~ \pi\,r^2 \;\cdot\;\frac{d\negmedspace h}{\dt} ~. \nonumber \]

The above equation forces you to consider only the derivative with respect to the time variable \(t\). What if you wanted to see the rates of change with respect to another variable, such as \(r\), \(h\), or some other quantity? In that case using the differential version of the above equation, namely

\[\dV ~=~ 2\pi \,rh\;\dr ~+~ \pi\,r^2 \;d\negmedspace h \nonumber \]

provides more flexibility—you are free to divide both sides by any differential, not just by \(\dt\). Many related rates problems would likely benefit from this approach.

Present-day calculus textbooks confuse the notion of a differential (infinitesimal) \(\dx\) with the idea of a small but real value \(\Delta x\). The two are not the same. An infinitesimal is not a real number and cannot be assigned a real value, no matter how small; \(\Delta x\) can be assigned real values. Using \(\dx\) and \(\Delta x\) interchangeably is a source of much confusion for students (likewise for \(\dy\) and \(\Delta y\)). This confusion rears its head in exercises involving the linear approximation of a curve by its tangent line near a point \(x_0\), namely \(f(x) \approx f(x_0) + f'(x_0)(x - x_0)\) when \(x-x_0\) is “small” (e.g. \(\sqrt{63} \approx 7.9375\), by using \(f(x)=\sqrt{x}\), \(x=63\), \(x_0=64\), and \(x-x_o = \Delta x = -1\)). Such exercises have nothing to do with differentials, not to mention having dubious value nowadays. They are remnants of a bygone era, before the advent of modern computing obviated the need for such (generally) poor approximations.

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2

Find the differential \(\df\) of \(f(x) = x^2 - 2x + 5\).

Find the differential \(\df\) of \(f(x) = \sin^2 (x^2)\).

2

Show that \(d\left(\tan^{-1}(y/x)\right) = \dfrac{x\,\dy ~-~ y\,\dx}{x^2 ~+~ y^2}\)

Given \(y^2 - xy + 2x^2 = 3\), find \(\dy\vphantom{\dfrac{x\,\dy ~-~ y\,\dx}{x^2 ~+~ y^2}}\).

The elasticity of a function \(y = f(x)\) is \(E(y) = \dfrac{x}{y} \cdot \dfrac{\dy}{\dx}~\). Show that \(E(y) = \dfrac{d(\ln\,y)}{d(\ln\,x)}~\).

Prove the differential version of the Quotient Rule:

\[d\left(\dfrac{f}{g}\right) ~=~ \dfrac{g\,\df ~-~ f\,\dg}{g^2} \nonumber \]

Let \(y = c\,u^n\), where \(c\) and \(n\) are constants. Show that

\[\frac{\dy}{y} ~=~ n\,\frac{\du}{u} ~. \nonumber \]

Obviously the derivative of the constant \(\pi^2\) is not \(2\pi\). But is \(d(\pi^2) = 2\pi\,d(\pi)\) true? Explain. [[1.]]

The continuity relation for an ideal gas is

\[\frac{PM}{\sqrt{T}} ~=~ \text{constant} \nonumber \]

where \(P\) and \(T\) are the pressure and temperature, respectively, of the gas, and \(M\) is the Mach number. Show that

\[\frac{\dP}{P} ~+~ \frac{d\!M}{M} ~=~ \frac{\dT}{2T} ~. \nonumber \]

For an ideal gas, satisfying the equation \(PV = RT\) as before, the Gibbs energy \(G\) is defined as \(G = H - TS\), where \(H\) and \(S\) are the enthalpy and entropy, respectively, of the gas.

1. Show that

   \[d\left(\dfrac{G}{RT}\right) ~=~ \dfrac{1}{RT}\,dG ~-~ \dfrac{G}{RT^2}\,\dT ~. \nonumber \]

2. One of the fundamental property relations for an ideal gas (which you do not need to prove) is

   \[dG ~=~ V\,\dP ~-~ S\,\dT ~. \nonumber \]

   Use this and part (a) to show that

   \[d\left(\dfrac{G}{RT}\right) ~=~ \dfrac{V}{RT}\,\dP ~-~ \dfrac{H}{RT^2}\,\dT ~. \nonumber \]

The derivative of the volume \(\pi r^2 h\) of a right circular cylinder of radius \(r\) and height \(h\), as a function of \(r\), equals its lateral surface area \(2\pi r h\). Use the notion of a differential as an infinitesimal change to explain why this makes sense geometrically.

The derivative of the volume \(\frac{4\pi}{3}r^3\) of a sphere of radius \(r\), as a function of \(r\), equals its surface area \(4\pi r^2\). Use the notion of a differential as an infinitesimal change to explain why this makes sense geometrically.

In quantum calculus the \(q\)-differential of a function \(f(x)\) is

\[d_qf(x) ~=~ f(qx) ~-~ f(x) ~, \nonumber \]

and the \(q\)-derivative of \(f(x)\) is

\[D_qf(x) ~=~ \frac{d_qf(x)}{d_qx} ~=~ \frac{f(qx) ~-~ f(x)}{qx ~-~ x} ~=~ \frac{f(qx) ~-~ f(x)}{(q - 1)x} ~. \nonumber \]

1. Show that for all positive integers \(n\),

   \[D_q\left(x^n\right) ~=~ \lbrack n \rbrack\, x^{n-1} ~, \nonumber \]

   where \(\lbrack n \rbrack = 1 + q + q^2 + \cdots + q^{n-1}\).

2. Use part (a) to show that for all positive integers \(n\),

   \[\lim_{q \to 1}~D_q\left(x^n\right) ~=~ \ddx\left(x^n\right) ~. \nonumber \]