Rules of differentiation

Differentiating from first principles — setting up $\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$ every single time — works, but it is slow and easy to fumble algebraically for anything beyond the simplest function. The rules of differentiation solve this once and for all: a short list of results, each provable from first principles, that can be applied mechanically to sums, differences, products, quotients, constant multiples and powers of known functions, plus the six basic trigonometric derivatives. This article collects those rules, proves the trickiest of them — the product rule — from the limit definition, and then works through the two situations that trip up most Class 11 students: a pure power-rule polynomial, and a genuine product that demands the product rule. The real skill this topic builds is not memorising formulas but reading an expression correctly before reaching for one.

The core insight

A small, provable set of rules — constant multiple, sum/difference, product, quotient and power — lets any combination of known differentiable functions be differentiated directly, without repeating the limit process each time.

The core idea

Let $u = u(x)$ and $v = v(x)$ be differentiable functions of $x$ , and let $c$ be a constant. Four rules handle every way two functions can be combined algebraically, and one rule handles powers.

Constant multiple and sum/difference. Scaling a function by a constant scales its derivative by the same constant, and the derivative of a sum or difference is the sum or difference of the derivatives.

\frac{d}{dx}\big[c\,u(x)\big] = c\,\frac{du}{dx}, \qquad \frac{d}{dx}\big[u(x)\pm v(x)\big] = \frac{du}{dx} \pm \frac{dv}{dx}

where: $u,v$ are differentiable functions of $x$ , and $c$ is a constant that does not depend on $x$ . These two rules are what let a polynomial be differentiated term by term.

Product rule. When two functions of $x$ are multiplied together, the derivative is not the product of the two derivatives — each factor is differentiated in turn while the other is held in place, and the two results are added.

\frac{d}{dx}\big[u(x)\,v(x)\big] = u(x)\,\frac{dv}{dx} + v(x)\,\frac{du}{dx}

where: $u\dfrac{dv}{dx}$ is "first function times derivative of second" and $v\dfrac{du}{dx}$ is "second function times derivative of first"; the two terms are added, never multiplied.

Quotient rule. For a ratio of two functions, with $v(x) \neq 0$ :

\frac{d}{dx}\left[\frac{u(x)}{v(x)}\right] = \frac{v(x)\dfrac{du}{dx} - u(x)\dfrac{dv}{dx}}{[v(x)]^2}

where: the numerator is "denominator times derivative of numerator, minus numerator times derivative of denominator," in that order, and the whole thing sits over the square of the denominator.

Power rule and basic trigonometric derivatives. For any real number $n$ ,

\frac{d}{dx}\big(x^n\big) = n\,x^{n-1}, \qquad \frac{d}{dx}(\sin x)=\cos x, \quad \frac{d}{dx}(\cos x)=-\sin x, \quad \frac{d}{dx}(\tan x)=\sec^2 x

with $\frac{d}{dx}(\cot x)=-\csc^2 x$ , $\frac{d}{dx}(\sec x)=\sec x\tan x$ , and $\frac{d}{dx}(\csc x)=-\csc x\cot x$ completing the set. The chain rule for composite functions like $\sin(2x)$ , derivatives of exponential/logarithmic/inverse-trigonometric functions, implicit differentiation and second-order derivatives belong to Class XII and are out of scope here. The figure below is the single most useful habit this topic builds: before differentiating anything, classify the expression as a sum of terms, a product of factors, or a quotient — the choice of rule follows immediately from that classification.

Figure

A decision flow showing an expression sorted into a sum branch handled by the power and sum rules, a product branch handled by the product rule, and a quotient branch handled by the quotient rule

Derivation AProving the product rule from first principles

The goal is to show, starting from nothing but the limit definition of a derivative and the ordinary algebra of adding and subtracting zero, that the derivative of a product $y = u(x)v(x)$ really does equal $u\,\frac{dv}{dx} + v\,\frac{du}{dx}$ , and not the tempting-but-wrong guess $\frac{du}{dx}\cdot\frac{dv}{dx}$ that a student who has only just met the sum rule might expect.

Step 1. Write the limit definition for $y = u(x)v(x)$ directly: $\dfrac{dy}{dx} = \lim_{h\to 0} \dfrac{u(x+h)v(x+h) - u(x)v(x)}{h}$ . This is the same starting point used for every derivative — the change in the function, divided by the change in $x$ , as that change shrinks to zero.

Step 2. Insert a term that changes nothing. Add and subtract $u(x+h)v(x)$ in the numerator: $u(x+h)v(x+h) - u(x+h)v(x) + u(x+h)v(x) - u(x)v(x)$ . This looks like an unmotivated trick, but it is exactly the move that lets the numerator be grouped into two pieces, each of which is recognisable.

Step 3. Group and factor. The first pair factors as $u(x+h)\big[v(x+h) - v(x)\big]$ , and the second pair factors as $v(x)\big[u(x+h) - u(x)\big]$ . So $\dfrac{dy}{dx} = \lim_{h\to 0}\left[ u(x+h)\cdot\dfrac{v(x+h)-v(x)}{h} + v(x)\cdot\dfrac{u(x+h)-u(x)}{h} \right]$ .

Step 4. Take the limit of each piece separately, using the algebra of limits (the limit of a sum is the sum of the limits, and the limit of a product is the product of the limits). As $h \to 0$ : $u(x+h) \to u(x)$ because $u$ is differentiable and hence continuous; $\dfrac{v(x+h)-v(x)}{h} \to \dfrac{dv}{dx}$ by definition; and $\dfrac{u(x+h)-u(x)}{h} \to \dfrac{du}{dx}$ by definition. Substituting gives $\dfrac{dy}{dx} = u(x)\dfrac{dv}{dx} + v(x)\dfrac{du}{dx}$ , exactly the product rule stated above — with no cross-term $\frac{du}{dx}\cdot\frac{dv}{dx}$ anywhere in sight.

Worked example

Differentiate $y = x^2\sin x$ with respect to $x$ .

Step 1. Identify the structure. This is a product of two functions, $u(x) = x^2$ and $v(x) = \sin x$ , so the product rule is needed — the power rule alone cannot handle it, because $x^2\sin x$ is not a single power of $x$ .

Step 2. Differentiate each factor separately: $\dfrac{du}{dx} = \dfrac{d}{dx}(x^2) = 2x$ and $\dfrac{dv}{dx} = \dfrac{d}{dx}(\sin x) = \cos x$ .

Step 3. Apply the product rule, $\dfrac{d}{dx}(uv) = u\dfrac{dv}{dx} + v\dfrac{du}{dx}$ : $\dfrac{dy}{dx} = x^2\cdot\cos x + \sin x\cdot 2x$ .

Step 4. Simplify and factor for the final form.

\frac{dy}{dx} = x^2\cos x + 2x\sin x = x\big(x\cos x + 2\sin x\big)

The derivative of $y = x^2\sin x$ is $\dfrac{dy}{dx} = x^2\cos x + 2x\sin x$ .

Competitive stretch

The same two functions, arranged as a quotient instead, $y = \dfrac{x^2}{\sin x}$ (with $\sin x \neq 0$ ), needs the quotient rule with the same $u = x^2$ , $v = \sin x$ , $\dfrac{du}{dx}=2x$ , $\dfrac{dv}{dx}=\cos x$ :

\frac{dy}{dx} = \frac{\sin x\cdot 2x - x^2\cdot\cos x}{\sin^2 x} = \frac{2x\sin x - x^2\cos x}{\sin^2 x}

Comparing this to the product-rule answer above shows how the same building blocks, combined differently, force a completely different rule and a different-looking answer — recognising the combination correctly is the whole game.

Common traps

Applying the power rule to a product or quotient of two functions of $x$ instead of the product or quotient rule — e.g. treating $x^2\sin x$ as if it were a single power and differentiating term by term.
Guessing that the derivative of a product is the product of the derivatives, $\dfrac{du}{dx}\cdot\dfrac{dv}{dx}$ — the first-principles derivation above shows explicitly why that term never appears.
Misremembering the quotient rule's order or sign — it is "denominator times derivative of numerator, minus numerator times derivative of denominator, all over denominator squared," not the reverse.
Forgetting that the derivative of a constant is $0$ , and mishandling constant terms inside a sum.
Mixing up the signs in the trigonometric derivatives, especially $\frac{d}{dx}(\cos x) = -\sin x$ versus $\frac{d}{dx}(\sin x) = \cos x$ .
Reaching for the chain rule on composite expressions like $\sin(2x)$ or $(x^2+1)^3$ — that rule is Class XII scope and is not required here.

Ask the tutor: not sure whether an expression needs the product rule, the quotient rule, or just the power rule term by term — ask your AI tutor to walk through classifying it with you.

Why can I not just multiply the derivatives of the two factors in a product

Because that is not what the limit definition gives. Differentiating $y=uv$ from first principles produces two terms, $u\frac{dv}{dx}$ and $v\frac{du}{dx}$ , added together — never a term with $\frac{du}{dx}\cdot\frac{dv}{dx}$ in it. You can see exactly where that cross-term would have to come from and why it does not survive the limit: only the terms where one factor is evaluated at $x$ (not $x+h$ ) remain after taking $h \to 0$ . Trust the derivation, not the shortcut.

How do I tell whether an expression is a product, a quotient, or just a sum of power terms

Look at how the pieces are joined. If terms are added or subtracted, like $4x^3 - 5x^2 + 7x$ , that is a sum — use the power rule term by term. If two genuinely different functions of $x$ are multiplied, like $x^2\sin x$ , that is a product — use the product rule. If one function of $x$ is divided by another, like $x^2/\sin x$ , that is a quotient — use the quotient rule. A single power of $x$ multiplied only by a constant, like $7x$ , is still just a power-rule term; the constant is not a "second function."

Do I need to memorise the product and quotient rule proofs for exams

Usually only the statement and application are tested at this level, but knowing the first-principles proof of the product rule is valuable insurance: it is exactly the kind of "derive this result" question that can appear, and it stops you from misremembering the formula under pressure, since you can rebuild it from the definition of a derivative if your memory slips.

Rules of differentiation

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads