UV Differentiation Formula — Product Rule Proof

TL;DR

The uv differentiation formula — also called the product rule — states that ( \frac{d}{dx}(uv) = u' v + u v' ). This article gives the formula, its first-principles proof, three worked examples at three difficulty tiers, the side-by-side trap with the false rule ( (uv)' = u' v' ), and the history of how Leibniz wrote down the rule in 1684.

A Rule That Says the Derivative of a Product Is Not the Product of Derivatives

Most students meet ( \frac{d}{dx}(x^2) = 2x ) in Class 11 and assume the derivative operation "passes through" multiplication. Then they try to differentiate ( x^2 \sin x ) as ( (2x)(\cos x) = 2x \cos x ) — and the answer is wrong.

The uv differentiation formula says:

[ \frac{d}{dx}(uv) = u'v + uv' ]

The derivative of a product is the first function times the derivative of the second, plus the second function times the derivative of the first. There are two terms — not one. That's the whole rule.

The Formula

For two differentiable functions ( u(x) ) and ( v(x) ):

[ \boxed{\frac{d}{dx}\bigl(u(x) \cdot v(x)\bigr) = u'(x) \cdot v(x) + u(x) \cdot v'(x)} ]

Shorter forms commonly used:

[(uv)' = u'v + uv' \quad\quad \frac{d(uv)}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} ]

Quick facts.

Type: A differentiation rule for products of differentiable functions.

Other names: the product rule, Leibniz's rule (for the first derivative).

Grade introduced: CCSS-M (AP Calculus, post-secondary) — derivative rules; NCERT Class 12 Chapter 5 — Continuity and Differentiability.

First written down: Gottfried Wilhelm Leibniz, 1684, in Acta Eruditorum — the first calculus paper ever published.

Related rules: quotient rule, chain rule, three-function product rule ( (uvw)' = u'vw + uv'w + uvw' ).

How the Formula Is Derived — First Principles

The cleanest proof comes from the definition of the derivative.

[ \frac{d}{dx}\bigl(u(x)v(x)\bigr) = \lim_{h \to 0} \frac{u(x+h)v(x+h) - u(x)v(x)}{h} ]

The numerator needs a trick — add and subtract ( u(x+h)v(x) ):

[ u(x+h)v(x+h) - u(x)v(x) = u(x+h)v(x+h) - u(x+h)v(x) + u(x+h)v(x) - u(x)v(x) ]

Group:

[ = u(x+h)[v(x+h) - v(x)] + v(x)[u(x+h) - u(x)] ]

Divide by ( h ):

[ \frac{u(x+h)v(x+h) - u(x)v(x)}{h} = u(x+h) \cdot \frac{v(x+h) - v(x)}{h} + v(x) \cdot \frac{u(x+h) - u(x)}{h} ]

Take the limit as ( h \to 0 ):

( u(x+h) \to u(x) ) (by continuity of ( u )).
( \frac{v(x+h) - v(x)}{h} \to v'(x) ) (definition of derivative).
( \frac{u(x+h) - u(x)}{h} \to u'(x) ) (definition of derivative).

What survives:

[ \frac{d}{dx}(uv) = u(x)v' + v(x)u' = uv' + u'v ]

The add-and-subtract trick is the entire proof. Once a student has seen ( u(x+h)v(x) - u(x+h)v(x) = 0 ) being inserted on purpose, the rule stops feeling mysterious.

Three Worked Examples — Quick, Standard, Stretch

Quick. Differentiate ( f(x) = x \cdot \sin x ).

Let ( u = x ) and ( v = \sin x ), so ( u' = 1 ) and ( v' = \cos x ).

[ f'(x) = u'v + uv' = (1)(\sin x) + (x)(\cos x) = \sin x + x \cos x ]

Final answer: ( \sin x + x \cos x ).

Standard (Wrong Path First — Watch How This Goes Wrong). Differentiate ( f(x) = x^2 e^x ).

The wrong path. A student multiplies the two derivatives directly: ( u = x^2 ), ( u' = 2x ); ( v = e^x ), ( v' = e^x ). The wrong claim — ( (uv)' = u' v' = (2x)(e^x) = 2x e^x ).

Check: differentiate at ( x = 1 ). The function is ( f(1) = 1 \cdot e = e \approx 2.718 ). The slope by the false rule is ( 2(1)(e) = 2e \approx 5.436 ) . The true slope is shown to be wrong.

The flaw: the derivative of a product is not the product of derivatives.

The rescue. Apply the formula:

[ f'(x) = (2x)(e^x) + (x^2)(e^x) = (e^x)(2x + x^2) = x(x + 2)e^x ]

Final answer: ( x(x + 2)e^x ).

Stretch. Differentiate ( f(x) = x^2 \sin x \cos x ).

Three functions multiplied. Apply the three-function product rule, or pair-and-group. Pair-and-group is cleaner: let ( u = x^2 ) and ( v = \sin x \cos x ). First compute ( v' ) using the product rule again:

[ v = \sin x \cos x \implies v' = \cos^2 x - \sin^2 x = \cos 2x ]

Now apply the formula:

[ f'(x) = (2x)(\sin x \cos x) + (x^2)(\cos 2x) ]

Using the identity ( 2 \sin x \cos x = \sin 2x ):

[ f'(x) = x \sin 2x + x^2 \cos 2x ]

Final answer: ( x \sin 2x + x^2 \cos 2x ).

Why the UV Differentiation Formula Matters — The Real-World Pay-off

The product rule isn't a textbook curiosity. It's the rule that lets calculus handle anything where two changing quantities multiply.

Physics — momentum and force. The product rule is used in determining momentum ( p = mv ) and force ( F = \frac{dp}{dt} = m'v + mv' ).
Economics — revenue. Revenue is price times quantity, differentiated with respect to time.
Quantum mechanics — expectation values. Time-evolution of expectation values requires multiple applications of the product rule.
Engineering — bending moments. Bending moment as a product of variable load distribution and moment arm.

The Mathematicians Behind the Product Rule

The product rule was published by Gottfried Wilhelm Leibniz in 1684 in Acta Eruditorum.

Callout — Leibniz and the first calculus paper.
When Leibniz published Nova Methodus pro Maximis et Minimis in October 1684 — just six pages — he did not yet know that Newton had derived the same rules nineteen years earlier.

Tripping Points to Avoid

Mistake 1: Multiplying the two derivatives directly.

Correct way: ( (uv)' = u'v + uv' ).

Mistake 2: Forgetting which function was differentiated.

Correct way: Write all four – ( u, u', v, v' ) – before assembling the answer.

Mistake 3: Using the product rule on composite functions.

Correct way: Use the chain rule instead.

The Short Version

The uv differentiation formula — the product rule — is ( (uv)' = u'v + uv' ).
The proof uses the add-and-subtract trick.
The first mistake common is writing ( (uv)' = u'v' ) — one term instead of two.
The rule was first published by Leibniz in 1684 in Acta Eruditorum.

Practice These Three Before Moving On

Differentiate ( f(x) = (3x + 1)(x^2 + 5) ).
Differentiate ( f(x) = x^3 \ln x ).
Differentiate ( f(x) = e^x \sin x \cos x ) (use product rule twice).