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:
$$= u(x+h)\bigl[v(x+h) - v(x)\bigr] + v(x)\bigl[u(x+h) - u(x)\bigr].$$
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:
$$\frac{d}{dx}(uv) = u(x) \cdot v'(x) + v(x) \cdot u'(x) = uv' + u'v.$$
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 $2(1)(e) = 2e \approx 5.436$. Numerical check on the actual function: $f(0.99) \approx 0.9801 \cdot e^{0.99} \approx 2.638$ and $f(1.01) \approx 1.0201 \cdot e^{1.01} \approx 2.802$. The slope $\approx (2.802 - 2.638)/0.02 \approx 8.2$. The false rule gives 5.4; the true slope is 8.2. The false rule is wrong by a clean factor.
The flaw: the derivative of a product is not the product of derivatives. The rule needs the cross-pattern $u'v + uv'$ — two terms, not one.
The rescue. Apply the formula:
$$f'(x) = u'v + uv' = (2x)(e^x) + (x^2)(e^x) = e^x(2x + x^2) = x(x + 2) e^x.$$
Check at $x = 1$: $f'(1) = 1 \cdot 3 \cdot e = 3e \approx 8.155$. That matches the numerical slope ≈ 8.2 within rounding.
Final answer: $x(x + 2)e^x$.
The lesson — the derivative of a product carries both directions of differentiation, not one.
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:
$$v = \sin x \cos x \implies v' = (\cos x)(\cos x) + (\sin x)(-\sin x) = \cos^2 x - \sin^2 x = \cos 2x.$$
Now apply the formula to $f = u v$:
$$f'(x) = u'v + uv' = (2x)(\sin x \cos x) + (x^2)(\cos 2x).$$
Using the identity $2 \sin x \cos x = \sin 2x$:
$$f'(x) = x \sin 2x + x^2 \cos 2x.$$
Final answer: $x \sin 2x + x^2 \cos 2x$.
The stretch example shows how the product rule applies recursively when more than two functions are multiplied — and how trig identities can collapse the answer to a cleaner form at the end.
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. If mass $m(t)$ and velocity $v(t)$ both change with time (a rocket burning fuel, say), then momentum $p = mv$ and force $F = dp/dt = m'v + mv'$. That's the product rule. Newton's "F = ma" is the special case where $m' = 0$.
Economics — revenue. Revenue is price × quantity. If both price $p(t)$ and quantity $q(t)$ change with time, marginal revenue is $R'(t) = p'q + pq'$. That decomposition is how economists distinguish price effects from quantity effects.
Quantum mechanics — expectation values. Time-evolution of the expectation value of an operator $\hat{A}$ involves $\frac{d}{dt}\langle\psi | \hat{A} | \psi\rangle$ — three product-rule applications, one for each factor.
Engineering — bending moments. In beam analysis, the bending moment is the product of a load distribution and a moment arm, both of which can vary along the beam. The product rule gives the rate at which the moment changes along the beam.
The Class 12 student who learns the product rule properly is already inside a Class-13/freshman-physics framework. The same rule that differentiates $x^2 \sin x$ tracks the momentum of a falling object losing mass.
The Mathematicians Behind the Product Rule
The product rule is one of the few formulas with a clean, documented origin — it was published by Gottfried Wilhelm Leibniz in 1684 in Acta Eruditorum, in the paper Nova Methodus pro Maximis et Minimis — the first calculus paper in history.
Gottfried Wilhelm Leibniz (1646–1716, Germany) developed his version of calculus independently of Isaac Newton (1643–1727, England). The notation we use today — $dy/dx$, the integral sign $\int$, the $d$ operator — is Leibniz's. So is the formal statement of the product rule.
Isaac Newton had derived the same rule independently in his Method of Fluxions (written 1671, published 1736 posthumously), in a different notation. The priority dispute between Newton and Leibniz over who invented calculus first lasted decades and divided British and Continental mathematics for over a century.
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 in unpublished work. The product rule appears on page 3 of Leibniz's paper, written in the form $d(xy) = x , dy + y , dx$ — the same formula Class 12 students learn today. Leibniz's notation won the long argument; Newton's results came first. The calculus the world uses is Leibniz's calculus in form and Newton's in spirit.
Tripping Points to Avoid
Mistake 1: Multiplying the two derivatives directly.
Where it slips in: A student differentiates $u \cdot v$ and writes $(uv)' = u' v'$ — one term instead of two.
Don't do this: Treat differentiation as a multiplicative operation that "passes through" a product.
The correct way: $(uv)' = u'v + uv'$. Two terms. Cross-pattern. The derivative of the first times the second, plus the first times the derivative of the second.
Mistake 2: Forgetting which function got differentiated.
Where it slips in: Writing $(x^2 \sin x)' = (2x)(\cos x) + (x^2)(\sin x)$ — second term uses the original $\sin x$ instead of its derivative $\cos x$. The cross-pattern got crossed too literally.
Don't do this: Apply the formula by memory without naming $u$, $v$, $u'$, $v'$ explicitly.
The correct way: Write down all four — $u$, $u'$, $v$, $v'$ — before assembling the answer. Then assemble.
On the Bhanzu McKinney whiteboard, the trainer draws a four-cell table — $u$, $u'$, $v$, $v'$ — and asks students to fill it in before the formula is written. Filling the table forces the second derivative computation, and the cross-pattern then snaps into place.
Mistake 3: Trying to use the product rule when the chain rule is the right tool.
Where it slips in: Differentiating $\sin(x^2)$, a student tries to write it as $u = \sin x$ and $v = x$ — but $\sin(x^2)$ isn't a product. It's a composition.
Don't do this: Force a product-rule decomposition on a composite function.
The correct way: $\sin(x^2)$ is $f(g(x))$ with $f(u) = \sin u$ and $g(x) = x^2$. Use the chain rule: $\frac{d}{dx}\sin(x^2) = \cos(x^2) \cdot 2x$. Product rule and chain rule are different tools.
The Short Version
The uv differentiation formula — the product rule — is $(uv)' = u'v + uv'$. Two terms, cross-pattern.
The proof uses the add-and-subtract trick $u(x+h)v(x+h) - u(x)v(x) = u(x+h)[v(x+h) - v(x)] + v(x)[u(x+h) - u(x)]$, divided by $h$, taken to the limit.
The single most common mistake is writing $(uv)' = u'v'$ — one term instead of two.
The rule was first published by Leibniz in 1684 in Acta Eruditorum — the founding paper of the calculus we use today.
Integration by parts — $\int u , dv = uv - \int v , du$ — is the integral version of this rule.
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).
If Problem 1 took you back to expanding the product first, that's a fine method too — but the product rule should give the same answer in two terms. If Problem 2 came out as $3x^2 \cdot \frac{1}{x}$, return to Mistake 1 above.
Want a live Bhanzu trainer to walk your child through the product, quotient, and chain rules with worked examples from JEE Main? Book a free demo class — online globally.
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