Differentials

Symbol	Meaning
$dx$	Independent differential of $x$; any real number
dy=f'(x)\,dx	Differential of $y$; depends on $x$ and $dx$
$\mathrm{\Delta }y$	True change in $y$: $f(x+\mathrm{\Delta }x)-f(x)$
$dy\approx \mathrm{\Delta }y$	When $dx=\mathrm{\Delta }x$ is small

Definition

Notes:

• $dx$ and $dy$ are single entities, not products of $d$ and $x$ (or $d$ and $y$).
• If $dx=0$, then $dy=0$.
• If $dx\ne 0$, dividing (i) by $dx$ gives dy/dx=f'(x), so the derivative equals the ordinary quotient of the differentials.
• $dx$ and $dy$ are not the limits of $\mathrm{\Delta }x$ and $\mathrm{\Delta }y$. Those limits are both zero.

Worked Examples

Geometric Interpretation

For fixed $x=x_0$, the equation dy=f'(x_0)\,dx defines a straight line through the origin in the $dx$-$dy$ plane, with slope f'(x_0).

If we place this $dx$-$dy$ plane on top of the $xy$-plane with the origin at $(x_0,f(x_0))$ and parallel axes, the line dy=f'(x_0)\,dx coincides with the tangent line to the curve at $(x_0,f(x_0))$.

Differentials as Approximations

When $\mathrm{\Delta }x=dx$ is small, the differential $dy$ approximates the true change $\mathrm{\Delta }y$:

$$\mathrm{\Delta }y\approx dy$$

or equivalently

\underbrace{f(x_0+dx)-f(x_0)}_{\Delta y}\approx\underbrace{f'(x_0)\,dx}_{dy}

Error Analysis

If the true value is $Q_t$ and the approximated value is $Q_a$:

$$\text{true error}=|Q_t-Q_a|=|\mathrm{\Delta }Q|$$$$\text{relative error}=\left|\frac{\mathrm{\Delta }Q}{Q_t}\right|\approx \left|\frac{dQ}{Q}\right|$$$$\text{percentage relative error}=\left|\frac{dQ}{Q}\right|\times 100$$

Properties of Differentials

Multiplying both sides of the derivative formulas by $dx$ yields the corresponding differential formulas:

Derivative Formula	Differential Formula
${\displaystyle \frac{d}{dx}}c=0$	$dc=0$
${\displaystyle \frac{d}{dx}}(cu)=c{\displaystyle \frac{du}{dx}}$	$d(cu)=cdu$
${\displaystyle \frac{d}{dx}}(u+v)={\displaystyle \frac{du}{dx}}+{\displaystyle \frac{dv}{dx}}$	$d(u+v)=du+dv$
${\displaystyle \frac{d}{dx}}(uv)=u{\displaystyle \frac{dv}{dx}}+v{\displaystyle \frac{du}{dx}}$	$d(uv)=udv+vdu$
${\displaystyle \frac{d}{dx}}\left({\displaystyle \frac{u}{v}}\right)={\displaystyle \frac{vdu/dx-udv/dx}{v^2}}$	$d\left({\displaystyle \frac{u}{v}}\right)={\displaystyle \frac{vdu-udv}{v^2}}$
${\displaystyle \frac{d}{dx}}(x^r)=r{x}^{r-1}$	$d(x^r)=r{x}^{r-1}dx$
${\displaystyle \frac{d}{dx}}\sin u=\cos u{\displaystyle \frac{du}{dx}}$	$d(\sin u)=\cos udu$
${\displaystyle \frac{d}{dx}}\tan u={\sec }^{2}u{\displaystyle \frac{du}{dx}}$	$d(\tan u)={\sec }^{2}udu$
${\displaystyle \frac{d}{du}}\arcsin u={\displaystyle \frac{1}{\sqrt{1-u^2}}}$	$d(\arcsin u)={\displaystyle \frac{du}{\sqrt{1-u^2}}}$

Invariance of the Form of the Differential

If $y=f(x)$ and $x$ is itself a function of another variable $t$ (so $x=g(t)$), then:

dy=f'(x)\,dx

holds whether $x$ is the independent variable or not. This is because if $x=g(t)$:

dy=h'(t)\,dt=f'(g(t))g'(t)\,dt=f'(x)\underbrace{g'(t)\,dt}_{dx}=f'(x)\,dx

This invariance means we can always write dy=f'(x)\,dx without caring whether $x$ is the base independent variable.

For example, for $y=\sin \sqrt{x}$, whether $x$ or $u=\sqrt{x}$ is independent:

$$dy=\cos \sqrt{x}\cdot \frac{1}{2\sqrt{x}}dx$$

Caution: Even though $dy/dx$ can be treated as a ratio when $x$ is the independent variable, we cannot "cancel" $dy$'s in the Chain Rule ${\displaystyle \frac{dz}{dx}}={\displaystyle \frac{dz}{dy}}{\displaystyle \frac{dy}{dx}}$ as an algebraic identity. The Chain Rule requires proof of existence and uses the full theorem, not just algebra.

Frequently Asked Questions

What is the difference between $\mathrm{\Delta }x$ and $dx$?

$\mathrm{\Delta }x$ is a concrete increment in $x$ that must keep $x_0+\mathrm{\Delta }x$ in the domain of $f$. The differential $dx$ is an independent variable ranging over all reals; it can take any value, not just small ones. When we use $dx=\mathrm{\Delta }x$ in approximations, we are treating $dx$ as small, but the definition does not require this.

Is $dy$ always a good approximation for $\mathrm{\Delta }y$?

Only when $dx=\mathrm{\Delta }x$ is small. The error $|\mathrm{\Delta }y-dy|$ is approximately \frac{1}{2}|f''(x_0)||(\Delta x)^2|, so the approximation improves as $\mathrm{\Delta }x\to 0$.

How do differentials apply to error propagation?

If a quantity $Q$ depends on a measured quantity $x$ via $Q=f(x)$, and $x$ has measurement error $\pm |\mathrm{\Delta }x|$, then the corresponding error in $Q$ is approximately |dQ|=|f'(x)||\Delta x|. The relative error is |dQ/Q|=|f'(x)/f(x)||\Delta x|.