Calculus · Differential Calculus
How much does the volume of a ball change if its radius grows by a millimetre? What is \(\sqrt{4.04}\) without a calculator? How accurate is a result computed from a measurement that has an error? All three questions have the same answer: use the differential. This post explains what it is, what it means on a graph, and how engineers and scientists use it every day.
Take a square with side \(x\). Its area is \(y = x^2\). If the side grows by a small amount \(\Delta x\), the area grows by
\[ \Delta y = (x + \Delta x)^2 - x^2 = \underbrace{2x\,\Delta x}_{\text{two thin strips}} + \underbrace{(\Delta x)^2}_{\text{tiny corner}}. \]Picture it: the new square is the old one plus two strips of size \(x \times \Delta x\) along two sides, plus a little corner square \(\Delta x \times \Delta x\). For \(x = 10\) and \(\Delta x = 0.01\), the strips add \(0.2\) and the corner only \(0.0001\).
The strips are proportional to \(\Delta x\): this is the linear part of the change. The corner is much smaller, it shrinks faster than \(\Delta x\) itself. For small changes we can ignore the corner and keep the simple linear part. That linear part is the differential.
Two things to notice:
For the independent variable itself, the differential is just its increment: \(dx = \Delta x\).
Divide the defining equation by \(\Delta x\):
\[ \frac{\Delta y}{\Delta x} = A(x) + \gamma \;\xrightarrow{\ \Delta x\to 0\ }\; A(x). \]So \(A(x)\) has to be the derivative! Conversely, if the derivative exists, set \( \gamma = \frac{\Delta y}{\Delta x} - f'(x) \), which tends to zero, and the decomposition holds with \(A = f'\).
This explains Leibniz's notation: dividing by \(dx\) gives \( \frac{dy}{dx} = f'(x) \). The derivative really is a quotient of two differentials. In older books, a function that has a differential at a point is called differentiable there, and by this theorem the word means the same thing as "has a finite derivative".
Since \( dy = f'(x)\,dx = \operatorname{tg}\alpha\cdot dx \), and \( \operatorname{tg}\alpha \) is the slope of the tangent, \(dy\) is how much the tangent rises when we move \(dx\) to the right.
So replacing \(\Delta y\) by \(dy\) means replacing the curve by its tangent near the point \(M\). Zoom in on any smooth curve far enough, and it looks like a straight line; the differential is the calculus of that straight line.
Since \(dy = y'\,dx\), every rule for derivatives becomes a rule for differentials, just multiply by \(dx\):
| Derivative rule | Differential rule |
|---|---|
| \( (c)' = 0 \) | \( dc = 0 \) |
| \( (cu)' = cu' \) | \( d(cu) = c\,du \) |
| \( (u \pm v)' = u' \pm v' \) | \( d(u \pm v) = du \pm dv \) |
| \( (uv)' = u'v + v'u \) | \( d(uv) = v\,du + u\,dv \) |
| \( \left(\frac uv\right)' = \frac{u'v - v'u}{v^2} \) | \( d\left(\frac uv\right) = \frac{v\,du - u\,dv}{v^2} \) |
Why: let \( y = f(u) \) with \( u = \varphi(x) \). By the chain rule, \( dy = f'(u)\,\varphi'(x)\,dx \). But \( \varphi'(x)\,dx = du \), so
\[ dy = f'(u)\,du, \]the same form as if \(u\) were independent. You can write differentials without first deciding which variable is "the" independent one. This is what makes substitutions in integrals (\(du = \varphi'(x)\,dx\)) work so smoothly.
Replacing the increment by the differential gives the most important practical formula of this topic:
The recipe: choose a "nice" point \(x\) close to where you need the value, where \(f(x)\) and \(f'(x)\) are easy to compute. Then correct by \(f'(x)\,\Delta x\).
Why \( f'(x) \ne 0 \)? The remainder \(\gamma\,\Delta x\) is small compared with \(\Delta x\). If \(f'(x) \ne 0\), then \(dy\) is of the same size as \(\Delta x\), so the remainder is also small compared with \(dy\), and \(dy\) is a good approximation in the relative sense. If \(f'(x) = 0\), then \(dy = 0\), and the whole change \(\Delta y\) sits in the "remainder"; the approximation tells us nothing useful.
Every measurement has an error. If we compute \(y = f(x)\) from a measured \(x\), how wrong can \(y\) be? The differential answers this.
The absolute error is measured in the same units as \(y\) ("give or take 3 cm³"). The relative error is a pure number, often given in percent ("accurate to 2%"). It's the absolute error divided by the value itself.
The relative error gets multiplied by the exponent. Cubes triple it (Example 6), square roots halve it. For a pendulum with period \( T = 2\pi\sqrt{l/g} \), a 1% error in the length \(l\) gives only a 0.5% error in the period.
| Concept | Formula |
|---|---|
| Increment decomposition | \( \Delta y = A(x)\,dx + \gamma\,dx,\ \ \gamma \to 0 \) |
| Differential | \( dy = f'(x)\,dx \) (exists \(\iff\) finite \(f'(x)\)) |
| Invariance of form | \( y = f(u),\ u = \varphi(x) \Rightarrow dy = f'(u)\,du \) |
| Geometric meaning | rise of the tangent over the step \(dx\) |
| Approximation | \( f(x + \Delta x) \approx f(x) + f'(x)\,\Delta x \) |
| Absolute error | \( |\Delta y| \approx |f'(x)\,\Delta x| \) |
| Relative error | \( \delta y \approx \left|\frac{f'(x)}{f(x)}\Delta x\right| \) |
Try these before opening the solution. More solved problems are in Differential Calculus.
\( dy = (3x^2 - 2)\,dx = 1\cdot 0.01 = 0.01 \). \( \Delta y = (1.01^3 - 2.02) - (1 - 2) = 1.030301 - 2.02 + 1 = 0.010301 \). The difference, \(0.000301\), is the small remainder.
\( dy = (\sin x + x\cos x)\,dx \).
\( x = 25 \), \( \Delta x = 1 \): \( \sqrt{26} \approx 5 + \frac{1}{10} = 5.1 \) (true value \(5.0990\dots\)).
\( \cos 61^\circ \approx \cos 60^\circ - \sin 60^\circ\cdot\frac{\pi}{180} \approx 0.5 - 0.8660\cdot 0.01745 \approx 0.4849 \) (true value \(0.48481\dots\)).
\( A = a^2 \), \( |\Delta A| \approx 2a\,|\Delta a| = 2\cdot 20\cdot 0.05 = 2 \) cm². Relative error: \( \delta A = 2\,\delta a = 2\cdot\frac{0.05}{20} = 0.005 = 0.5\% \).
\( d\left(\frac uv\right) = \left(\frac uv\right)'dx = \frac{u'v - v'u}{v^2}\,dx = \frac{v\,(u'dx) - u\,(v'dx)}{v^2} = \frac{v\,du - u\,dv}{v^2} \).
\( dA = 2\pi r\,dr = 2\pi\cdot 5\cdot 0.02 = 0.2\pi \approx 0.63 \) m². (This is the circumference times the thickness of the added ring.)
The differential captures one simple, powerful idea: for small changes, every smooth function behaves like a linear one. That idea lets you estimate values without a calculator, predict how measurement errors spread through a formula, and later, in integral calculus and differential equations, it becomes the basic language for writing "a tiny piece of something".
Next in the series: higher-order derivatives and differentials.