If you know how $u$ changes and how $v$ changes, how does the product $uv$ change? The natural guess is to multiply the changes: $d(uv)=du\cdot dv$. It looks so symmetrical, so elegant…

Leibniz made the same guess.

Leibniz tests and fails

In the same manuscript of 11 November 1675 in which he began using the letter $d$, Leibniz asked himself whether $d(xy)$ would equal $dx\,dy$, and whether $d\!\left(\frac xy\right)$ would equal $\dfrac{dx}{dy}$. Instead of trusting elegance, he tested it with an example and found it did not. A few days later he found the right rule.

That is an important lesson: even the inventors of calculus made mistakes, and what saved them was checking.

Try the test yourself. With $u=x$ and $v=x$, the product is $x^2$, whose derivative is $2x$. But the product of the derivatives is $1\cdot1=1$. It does not match.

The growing rectangle

Think of $u$ and $v$ as the sides of a rectangle, and $uv$ as its area. Now increase both sides a tiny bit: $u$ becomes $u+du$ and $v$ becomes $v+dv$.

The new area is $(u+du)(v+dv)$. What did it gain?

$$d(uv)=(u+du)(v+dv)-uv=v\,du+u\,dv+du\,dv.$$

Geometrically, the rectangle gained three pieces: a long strip of area $v\,du$, another long strip of area $u\,dv$, and a tiny square in the corner of area $du\,dv$.

The tiny square is the product of two infinitely small quantities, and so infinitely smaller than the strips. Leibniz neglected it and was left with

$$d(uv)=u\,dv+v\,du.$$

The guess $du\,dv$ was precisely the piece that does not matter.

The version with limits

Today we write $p(x)=f(x)\,g(x)$ and divide the gain by an increment $h\ne0$:

$$\frac{\Delta p}{h}=g\,\frac{\Delta f}{h}+f\,\frac{\Delta g}{h}+\frac{\Delta f}{h}\,\Delta g.$$

As $h\to0$, the first two terms tend to $gf'$ and $fg'$. The last tends to zero, because $\dfrac{\Delta f}{h}$ approaches $f'$, a fixed number, while $\Delta g$ goes to zero. Therefore,

$$(fg)'=f'g+fg'.$$

Example

For $f=x^2$ and $g=x+1$:

$$(fg)'=2x(x+1)+x^2\cdot1=3x^2+2x.$$

Check it another way: $x^2(x+1)=x^3+x^2$, whose derivative is $3x^2+2x$. At $x=1$ the product is $2$ and the derivative is $5$. The term $f'g$ contributes $4$ and the term $fg'$ contributes $1$.

f(x) = x²; g(x) = x + 1; p(x) = f(x)g(x).

Change x and compare the terms f′g and fg′. Their sum is the derivative of the product; the product of the derivatives is something else.

Leibniz’s rule

Leibniz went further and found how to differentiate a product several times. For the second derivative:

$$(uv)''=u''v+2u'v'+uv''.$$

The coefficients $1,2,1$ are the same as in $(a+b)^2$. For the third derivative we get $1,3,3,1$, as in $(a+b)^3$. In general the coefficients are the binomial ones, the same as in Pascal’s triangle. This formula is still called Leibniz’s rule.

Exercise: differentiate x(x² + 1)

With $f=x$ and $g=x^2+1$: $(fg)'=1\cdot(x^2+1)+x\cdot2x=3x^2+1$. Expanding first, $x^3+x$ has the same derivative.

Exercise: a rectangle has sides u = 3 cm and v = 5 cm. If u grows by 0.1 cm and v by 0.2 cm, how much does the area grow?

By the rule, $d(uv)\approx v\,du+u\,dv=5\cdot0.1+3\cdot0.2=1.1$ cm². The exact value is $3.1\cdot5.2-15=1.12$; the difference, $0.02$, is the tiny square $du\,dv=0.1\cdot0.2$.

Sources: Leibniz and calculus, MacTutor.

All Leibniz topics · ← Leibniz’s notation: a language for change · The ∫ sign: from sum to accumulation →

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