Why the derivative of a product is not the product of the derivatives, what the correct rule is, how a growing rectangle explains it, how to prove it, and how it extends to three or more factors.
A tempting guess is that the derivative of fg is f′g′. It fails at once, because x⋅x=x2 has derivative 2x while 1⋅1=1. The correct rule has two terms, one for each factor changing while the other holds still, and a picture of a growing rectangle makes it look obvious.
Key terms
Product rule
Definition If f and g are both differentiable at x, then fg is differentiable at x and
(fg)′(x)=f′(x)g(x)+f(x)g′(x).
In Leibniz notation, with u=f(x) and v=g(x), this reads
dxd(uv)=dxduv+udxdv.
In plain words Differentiate the first factor and keep the second, then keep the first and differentiate the second, and add.
Example Let y=(x2+1)(x3−2x). Then
y′=2x(x3−2x)+(x2+1)(3x2−2)=5x4−3x2−2.
Multiplying out first gives y=x5−x3−2x and the same derivative, which is a good check.
Geometric explanation
Definition Let u=f(x) and v=g(x) be the sides of a rectangle with area uv. When x changes to x+h, write Δu=f(x+h)−f(x) and Δv=g(x+h)−g(x). Then exactly
(u+Δu)(v+Δv)−uv=uΔv+vΔu+ΔuΔv.
The change in area is two thin strips, uΔv along one side and vΔu along the other, plus a small corner ΔuΔv. After dividing by h, the corner term is Δu⋅hΔv, which tends to 0⋅g′(x)=0.
In plain words When both sides of a rectangle grow a little, the new area is mostly two strips. The corner piece is so small it does not count in the limit.
Example Take u=3, v=5, Δu=0.1 and Δv=0.2. The new area is 3.1⋅5.2=16.12, so the area grew by 1.12. The strips give 3⋅0.2+5⋅0.1=1.1 and the corner gives 0.1⋅0.2=0.02, which add to 1.12.
Proof from difference quotients
Definition Add and subtract f(x+h)g(x) in the numerator.
As h→0 the second difference quotient tends to f′(x) and the first to g′(x). Also f(x+h)→f(x), because f is differentiable and therefore continuous at x. By the limit laws the whole expression tends to f(x)g′(x)+g(x)f′(x).
In plain words One small algebra trick splits the change in the product into the change caused by g and the change caused by f. Each piece becomes a familiar derivative.
Example Take f(x)=x2 and g(x)=x3 at x=1, so fg=x5 and the derivative should be 5. The two pieces of the split for shrinking h are shown below, rounded to four decimal places.
h
f(1+h)⋅hg(1+h)−g(1)
g(1)⋅hf(1+h)−f(1)
Sum
0.1
4.0051
2.1
6.1051
0.01
3.0910
2.01
5.1010
0.001
3.0090
2.001
5.0100
The pieces approach f(1)g′(1)=3 and g(1)f′(1)=2, and the sum approaches 5.
Products of several functions
Definition If f, g and k are differentiable at x, then applying the product rule twice gives
(fgk)′=f′gk+fg′k+fgk′.
In general, the derivative of a product of n differentiable functions is a sum of n terms, where each term differentiates one factor and leaves the others unchanged. In particular, for a positive integer n and f differentiable at x,
dxd(f(x)n)=nf(x)n−1f′(x).
In plain words Take turns. Each factor gets differentiated once while the rest stay as they are, and you add up all the turns.
Example For y=x(x+1)(x+2) at x=1, the three terms are 1⋅2⋅3=6, 1⋅1⋅3=3 and 1⋅2⋅1=2, so y′(1)=11. Expanding, y=x3+3x2+2x and y′=3x2+6x+2, which is also 11 at x=1. For a power, the last formula gives
dxd((x2+1)3)=3(x2+1)2⋅2x=6x(x2+1)2.
Check yourself
Differentiate y=(2x−1)(x2+4) with the product rule.
Show the answer
The rule gives
y′=2(x2+4)+(2x−1)(2x)=6x2−2x+8.
Expanding first gives y=2x3−x2+8x−4 and the same answer.
Suppose f(2)=3, f′(2)=−1, g(2)=4 and g′(2)=5. Find (fg)′(2).
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