# 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](https://mathshistory.st-andrews.ac.uk/Biographies/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 1. Differentiate \( f(x) = (3x + 1)(x^2 + 5) \). 2. Differentiate \( f(x) = x^3 \ln x \). 3. Differentiate \( f(x) = e^x \sin x \cos x \) (use product rule twice).