Differentials

A rise over a run

Between two points on a curve y = f(x), the run is the change in x, written $\Delta x$Δx, and the rise is the change in y, written $\Delta y$Δy. The gradient of the chord joining them is $\Delta y/\Delta x$Δy / Δx, a genuine fraction.

On $y=x^2$y = x² starting from x = 2, a run of 1 takes y from 4 to 9, so the chord's gradient is 5 ÷ 1 = 5. A run of 0.1 takes y from 4 to 4.41, a gradient of 0.41 ÷ 0.1 = 4.1. As the run shrinks, the gradient closes in on 4, and that limit is the derivative, written $\frac{dy}{dx}$dy/dx. The notation keeps the shape of the fraction it came from.

The curve $y=x^2$y = x² with a chord from x = 2 to x = 2 + h. At h = 1 the chord rises 5 over a run of 1, so its gradient is 5. Drag h down to 0 and the chord turns into the tangent at x = 2, whose gradient is 4.

Splitting $\frac{dy}{dx}$dy/dx into two pieces

The pieces dy and dx can be given meanings of their own. Let dx be a change in x, any size. Then define dy as the derivative times that change: dy = f'(x) dx. With these meanings, dividing dy by dx gives f'(x) exactly, so $\frac{dy}{dx}$dy/dx really is a ratio of two amounts.

For $y=x^2$y = x² the derivative is 2x, so dy = 2x dx. This is the rise along the tangent line: a line with gradient 2x that runs dx rises 2x times dx.

dy against the real change

The curve itself rises by $\Delta y={\left(x+\Delta x\right)}^2-x^2=2x\Delta x+{\left(\Delta x\right)}^2$Δy = (x + Δx)² − x² = 2xΔx + (Δx)². The first part is dy. The second part, ${\left(\Delta x\right)}^2$(Δx)², is what the curve adds by bending away from its tangent.

At x = 3 with a run of 0.1, dy = 2 × 3 × 0.1 = 0.6, while the curve rises $3.1^2-9=0.61$3.1² − 9 = 0.61. With a run of 0.01, dy = 0.06 and the curve rises $3.01^2-9=0.0601$3.01² − 9 = 0.0601. Making the run 10 times smaller made the leftover 100 times smaller, from 0.01 to 0.0001. So for a small run, dy is a very good estimate of the real change, and it gets better faster than the run shrinks.

The curve $y=x^2$y = x² and its tangent y = 2x − 1 at (1, 1). Over a run of 0.5, the tangent rises dy = 2 × 1 × 0.5 = 1, to (1.5, 2). The curve rises 1.25, to (1.5, 2.25). The gap between them is 0.25, which is $0.5^2$0.5².

Any letter works

The rule is the same whatever the letters are called: differentiate, then attach the change in the variable. If $u=x^3+1$u = x³ + 1, then $\frac{du}{dx}=3x^2$du/dx = 3x², so $du=3x^{2}dx$du = 3x² dx. If $s=t^2$s = t², then ds = 2t dt.

Check du against the real change at x = 2. With a run of 0.1, du = 3 × 4 × 0.1 = 1.2, while u changes from 9 to $2.1^3+1=10.261$2.1³ + 1 = 10.261, a change of 1.261. With a run of 0.01, du = 0.12 and the real change is 0.120601.

This is what integration by substitution, a later lesson, relies on. In $\int 3x^2{\left(x^3+1\right)}^{4}dx$∫ 3x²(x³ + 1)⁴ dx, the factor $3x^2$3x² and the dx together are exactly du, so the whole integral can be rewritten in terms of u.

Why the pieces may be separated

$\frac{dy}{dx}$dy/dx was defined as a limit of fractions, not as a fraction, so separating it needs a reason. The reason is the definition above: dy is chosen to be f'(x) dx, which makes the ratio of dy to dx equal to the gradient for every size of dx.

What links dy to the real change is that the two differ by an amount that shrinks faster than dx. At x = 3 on $y=x^2$y = x², that difference divided by the run was 0.1 for a run of 0.1 and 0.01 for a run of 0.01, heading to 0. As dy and dx shrink together, their ratio stays the gradient, and dy and the real rise become indistinguishable.

An estimate

A differential gives quick estimates. For $y=\sqrt{x}$y = √x the derivative is $\frac{1}{2\sqrt{x}}$1/(2√x), so $dy=\frac{dx}{2\sqrt{x}}$dy = dx/(2√x). At x = 4, where $\sqrt{4}=2$√4 = 2, a change of dx = 0.1 gives $dy=\frac{0.1}{4}=0.025$dy = 0.1/4 = 0.025, so $\sqrt{4.1}$√4.1 is about 2 + 0.025 = 2.025. The true value is 2.02485, so the estimate is off by about 0.00015.

The usual mistakes

Leaving off the dx. For $u=x^3+1$u = x³ + 1, du is $3x^{2}dx$3x² dx, not $3x^2$3x². The gradient alone is a rate; a differential is an amount.

Using the function instead of its derivative. For $y=x^2$y = x², dy is 2x dx, not $x^{2}dx$x² dx.

Treating dy as the exact change. dy is the change along the tangent; the curve's change differs from it by a small amount, 0.01 at x = 3 with a run of 0.1.

Replacing only the dx in a substitution. In $\int 2x{\left(x^2+5\right)}^{3}dx$∫ 2x(x² + 5)³ dx with $u=x^2+5$u = x² + 5, du = 2x dx replaces the 2x and the dx together.