Product rule : Product rule (calculus) 

Common misconception

It is a common misconception, when studying calculus, to suppose that the derivative of (uv) equals (u')(v'); however, it is quite easy to find counterexamples to this. Most simply, take a function f, whose derivative is f '(x). Now that function can be also written f(x) · 1, since 1 is the identity element for multiplication. Now, suppose the above mentioned misconception were true; if so, (u')(v') would equal zero; since, the derivative of a constant (such as 1) is zero; and, the product, of any number and zero, is zero. Such a misconception could result in a belief that the derivative, of every function, is 0; such a belief isn't correct.

Informal Proof

Informal justification of the product rule

The product rule can be justified as follows. Let:

<math>y = uv</math>

and suppose that you want to find dy/dx.

Let the independent variable x increase by a small amount Δx, resulting in an increase of u by Δu, of v by Δv, and of y by Δy. Now:

<math>y + \Delta y = (u + \Delta u)(v + \Delta v)</math>

Multiply out to obtain:

<math>y + \Delta y = uv + u \Delta v + v \Delta u + \Delta u \Delta v = y + u \Delta v + v \Delta u + \Delta u \Delta v</math>

So, cancelling y from the left and right hand sides,

<math>\Delta y = u \Delta v + v \Delta u + \Delta u \Delta v</math>

Now divide throughout by Δx:

<math>\frac{\Delta y}{\Delta x} = u \frac{\Delta v}{\Delta x} + v \frac{\Delta u}{\Delta x} + \Delta u \frac{\Delta v}{\Delta x}</math>

Let Δx tend to 0, then Δu tends to 0 also, since u is a differentiable and hence continuous function of x. We get:

<math>\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx} + 0\frac{dv}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}</math>

which proves the product rule.

Generalizations

The product rule can be generalised to products of more than two factors. For example, for three factors we have

<math>\frac{d(uvw)}{dx} = \frac{du}{dx}vw + u\frac{dv}{dx}w + uv\frac{dw}{dx}</math>

It can also be generalized to higher derivatives of products of two factors: if y = uv and y⁽ⁿ⁾ denotes the n-th derivative of y, then

<math>y^{(n)}(x) = \sum_{k=0}^n {n \choose k} u^{(n-k)}(x)\; v^{(k)}(x)</math>

(see binomial coefficient). This result is formally quite similar to the binomial theorem.

In multivariable calculus, the product rule is also valid for different notions of "product": scalar product and cross product of vectors, matrix product, inner products etc. All of these are summarized by the following general statement: let X, Y, Z be Banach spaces (which includes Euclidean space) and let B : X × Y → Z be a continuous bilinear operator. Then B is differentiable, and its derivative at the point (x,y) in X × Y is the linear map D_(x,y)B : X × Y → Z given by

D_(x,y)B(u,v) = B(x,v) + B(u,y)   for all (u,v) in X × Y.

See also:

• Quotient Rule

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