Differentiation rules and the chain rule

Two changing factors make one changing product

A rectangle has side lengths x and x+1, in metres, with x>0. Its area is P(x)=x(x+1) square metres. At x=2, the area is 6. Increasing x by a nonzero h gives P(2+h)−P(2)=5h+h², so the area change divided by the input change is 5+h square metres per metre.

As h approaches zero, this rate approaches 5. Both side lengths change at rate 1, but multiplying those two rates would give 1, not 5. One changing side acts on the other side’s current length; there are two contributions to account for.

A derivative is this finite limiting rate at a specified input. The records below compare exact quotients for (x+1)(2x+1), another product with different factor rates. Predict its derivative at the supplied input before revealing the result. A few records check arithmetic; the identity and proof explain every nearby input.

The limit needed by every rule

Prepare by substituting into expressions, factoring polynomials, and following a function composition. The linked lessons practise these already available skills. First read the concrete product, quotient and composition cases. Return to the error estimates and real-power proof when you can follow their inequalities; the derivative notation and needed limit facts are recalled here.

Let f be a real-valued function defined on an open interval around a, including a. For h≠0 sufficiently small, put Q_f(h)=[f(a+h)−f(a)]/h. The finite derivative f′(a)=A means: for every ε>0 there is δ>0 such that 0<|h|<δ implies |Q_f(h)−A|<ε. The radius also keeps a+h inside that interval. The prime and d/dx denote this limit, rather than an ordinary division by zero.

Here |v| is the absolute value, the distance from v to zero. We use |u+v|≤|u|+|v| and |uv|=|u||v|. A function is continuous at a if f(a+h) approaches the actual value f(a). Differentiability implies this continuity: for small h, |Q_f(h)|≤|A|+1, so |f(a+h)−f(a)|≤(|A|+1)|h|. For a requested positive tolerance choose an input radius smaller than that tolerance divided by |A|+1.

We will also use elementary finite-limit arithmetic. If U→u and V→v, sums converge by sharing error budgets. For products, UV−uv=V(U−u)+u(V−v); keep |V|≤|v|+1, then request errors smaller than ε/[2(|v|+1)] and ε/[2(|u|+1)]. Their two contributions total less than ε, with a zero contribution harmless. If v≠0, |V−v|<|v|/2 ensures |V|>|v|/2 and |1/V−1/v|≤2|V−v|/|v|², proving the reciprocal limit. These estimates justify the finite-limit operations below.

Constants and sums preserve the difference quotient

For a constant f(x)=c, Q_f(h)=0, so f′(a)=0. For f(x)=x, the quotient is 1, so f′(a)=1. A fixed real number k gives (kx+b)′=k because [k(a+h)+b−(ka+b)]/h=k.

If f and g are differentiable at the same input a on a shared open interval, Q_(f+g)=Q_f+Q_g and Q_(f−g)=Q_f−Q_g. The finite-limit sum and difference estimates give (f±g)′(a)=f′(a)±g′(a). For a fixed real c, Q_(cf)=cQ_f, hence (cf)′(a)=cf′(a); if c=0 both sides are zero.

Fixed means independent of x. In c x², c stays the same when x changes. In x·x², the first factor also changes, so the fixed-scalar rule cannot treat it as c. Nor can a sum rule be reversed: |x| and −|x| are not differentiable at zero, while their sum is the differentiable constant zero.

Split the product change into two contributions

Assume f and g have finite derivatives at a on a shared open interval. Write f₀=f(a), g₀=g(a), Δf=f(a+h)−f₀ and Δg=g(a+h)−g₀. Expanding gives Δ(fg)=g₀Δf+f₀Δg+ΔfΔg. The two thin strips and corner of the rectangle below show this identity for positive lengths and positive increments; the algebra also applies to signed values.

Divide by h≠0. Then Q_(fg)(h)=g₀Q_f(h)+f₀Q_g(h)+Δf Q_g(h). The first two terms approach g₀f′(a) and f₀g′(a). Continuity gives Δf→0, and the finite derivative keeps Q_g bounded, so the last term approaches zero. Therefore (fg)′(a)=f′(a)g(a)+f(a)g′(a).

For f=x+1 and g=2x+1, the rule gives P′(a)=1·(2a+1)+(a+1)·2=4a+3. Expanding P=2x²+3x+1 independently gives Q_P(h)=4a+3+2h. At a=1 the derivative is 7. In the area diagram h=1/4, Δf=1/4 and Δg=1/2; the corner contributes ΔfΔg/h=1/2 to that finite quotient, and tends to zero as h shrinks.

• Original area
• Strip from the second side’s change
• Strip from the first side’s change
• Corner from both changes

The corner term divided by h is 1/2 in this supplied finite case; for these linear factors it equals 2h and approaches zero. This area picture requires positive side lengths and positive increments; the product identity also covers signed changes.

A reciprocal needs a nonzero value

If g is differentiable at a and g(a)=g₀≠0, continuity supplies a whole nearby interval with |g(a+h)−g₀|<|g₀|/2. Thus |g(a+h)|>|g₀|/2>0. The reciprocal is actually defined at a and throughout this smaller interval.

Its exact quotient is Q_(1/g)(h)=−Q_g(h)/[g(a+h)g₀]. Finite-limit arithmetic gives (1/g)′(a)=−g′(a)/g₀². Applying the product rule to f·(1/g) proves (f/g)′(a)=[f′(a)g(a)−f(a)g′(a)]/[g(a)]². Keep the numerator order: derivative of the numerator times the denominator, minus the numerator times derivative of the denominator.

For R(x)=(x+1)/(x−1), the real domain excludes 1. At any a≠1, R′(a)=[(a−1)−(a+1)]/(a−1)²=−2/(a−1)². At a=2 it is −2. At a=1 the original formula has no point value, so it has no derivative there. A rule failure alone is not a universal verdict: x²/x on x≠0 can be extended by assigning value zero at zero, making a new function equal to x and differentiable there.

Build the power rule from products

For every positive integer n, (xⁿ)′=n xⁿ⁻¹ on the whole real line. The base case n=1 is the derivative of x, equal to 1, including at zero. If the formula holds for n, the product rule applied to x·xⁿ gives xⁿ+x·n xⁿ⁻¹=(n+1)xⁿ, proving the next case. This induction establishes every positive integer exponent, rather than extrapolating from a list.

The constant function P₀(x)=1 separately has derivative zero everywhere; this definition does not settle the undefined expression 0⁰. If n=−m with a positive integer m, define xⁿ=1/xᵐ only for x≠0. The reciprocal rule gives −m xᵐ⁻¹/x²ᵐ=n xⁿ⁻¹ on that domain. The derivative formula never restores an excluded input zero.

A polynomial is a finite sum of fixed real multiples of nonnegative integer powers, with its constant term interpreted as a constant function. Hence 3x⁴−2x+5 has derivative 12x³−2 at every real input. A rational function is differentiable wherever its polynomial denominator is nonzero. Record that original domain even after simplifying its derivative.

Follow the inner input before the outer rate

In a composition, u=f(x) becomes the input to y=g(u); write y=(g∘f)(x)=g(f(x)). Suppose f is differentiable at a and g at b=f(a), with real open intervals around a and b and all sufficiently nearby inner values in g’s domain. The chain rule says (g∘f)′(a)=g′(f(a))f′(a). Evaluate the outer derivative at the inner value, not automatically at a.

For f(x)=2x and g(u)=u², the composition is 4x². Its derivative is g′(2a)·2=(4a)·2=8a. At a=1 the intermediate input is 2, the outer rate is 4, and the composite rate is 8. Omitting the inner rate would give 4; evaluating g′ at a would also be wrong.

For H(x)=(x²+1)², use inner f=x² and outer g(u)=(u+1)². The derivative is 2(x²+1)·2x=4x(x²+1). With more layers, repeat this input tracking: [(3x+1)²+2]³ has derivative 3[(3x+1)²+2]²·2(3x+1)·3. Every factor corresponds to a layer, evaluated at that layer’s actual input.

Prove the chain rule without dividing by a zero inner change

Let A=f′(a), b=f(a) and B=g′(b). For a small allowed outer increment k≠0 define e(k)=[g(b+k)−g(b)]/k−B, and set e(0)=0. Differentiability makes e(k)→0 as k→0, including its assigned zero value. The identity g(b+k)−g(b)=k[B+e(k)] now holds for every small k, including k=0.

For h≠0 put k(h)=f(a+h)−f(a). Continuity of f gives k(h)→0, and k(h)/h=Q_f(h)→A. Consequently Q_(g∘f)(h)=Q_f(h)[B+e(k(h))]. This identity does not require dividing by k(h), which might be zero for some or all nearby inputs.

Here is the complete error control. Restrict to |Q_f(h)|≤|A|+1. Given ε>0, request |Q_f(h)−A|<ε/[2(|B|+1)]. Choose an outer radius η>0 making |e(k)|<ε/[2(|A|+1)] for |k|<η; at k=0 this is true because e(0)=0. Continuity of f supplies an input radius making |k(h)|<η. Use the minimum of these radii and the domain radii. Then |Q_(g∘f)−AB|≤|B||Q_f−A|+|Q_f||e(k(h))|<ε. If B=0 the first term is zero; both budgets remain positive. This proves the rule, including A=0 and every zero inner increment.

For f(x)=0 when x≤0 and f(x)=x² when x>0, f′(0)=0: its left quotient is zero and its right quotient is h. Compose with g(u)=(u+1)². The composite quotient is zero for h<0 and 2h+h³ for h>0, so its derivative is zero. On the entire left side k(h)=0; the proof still works, whereas a shortcut dividing by k(h) would fail.

Prove root and rational-power derivatives on positive inputs

For a positive integer q and x>0, x¹⁄q is the unique positive qth root. At a>0 let b=a¹⁄q. The root approaches b as x→a: for any small η with 0<η<b, the two numbers (b−η)ᑫ and (b+η)ᑫ lie strictly below and above a. Keeping x between them forces its positive root between b−η and b+η, by strict increase of positive integer powers.

Put t=x¹⁄q. For x≠a, the factorisation x−a=tᑫ−bᑫ=(t−b)(tᑫ⁻¹+tᑫ⁻²b+⋯+bᑫ⁻¹) gives (t−b)/(x−a)=1/(tᑫ⁻¹+tᑫ⁻²b+⋯+bᑫ⁻¹). The denominator has q positive terms and approaches q bᑫ⁻¹≠0. Thus (x¹⁄q)′ at a is 1/(q a^(1−1/q)). The q=1 case has a single term equal to 1.

For a rational exponent r=p/q with integer p and positive q, xʳ=(x¹⁄q)ᵖ on x>0. The integer power and chain rules give p bᵖ⁻¹/[q bᑫ⁻¹]=(p/q)a^(p/q−1). For p=0 use the constant function directly. Thus (xʳ)′=r xʳ⁻¹ on positive inputs, including negative rational exponents without permitting x=0.

Extend the fixed-exponent rule to real powers

For x>0 and a fixed real exponent r, xʳ denotes the usual continuous extension of rational powers in the exponent: approximate r by rational exponents. These positive powers satisfy (xy)ʳ=xʳyʳ, xʳ⁻¹=xʳ/x and 1ʳ=1. For y>1 they increase with the exponent; for 0<y<1 they decrease. These are algebraic power properties, not derivative formulas: rational powers have them by roots and integer products, and passing to the defining continuous extension preserves them.

At the input 1 choose rational s<r<t. Such choices can be made as close as desired: for any length ℓ>0, choose an integer denominator q with 1/q<ℓ and the multiples of 1/q immediately below and above r. For y>1, yˢ≤yʳ≤yᵗ. For 0<y<1 the power inequalities reverse, but division by the negative y−1 reverses them again. On both sides, (yˢ−1)/(y−1)≤(yʳ−1)/(y−1)≤(yᵗ−1)/(y−1).

The two outer quotients approach s and t by the proved rational rule. Given ε>0 choose rationals with r−ε/2<s<r<t<r+ε/2, then restrict |y−1| so the outer quotient errors are less than ε/2. The middle quotient lies strictly between r−ε and r+ε. Hence its limit is r: (xʳ)′ at x=1 is r, for every fixed real r.

For any a>0 set y=x/a. Then [xʳ−aʳ]/(x−a)=aʳ⁻¹[(yʳ−1)/(y−1)]. As x→a, y→1, so (xʳ)′=r aʳ⁻¹. The exponent must stay fixed. This rule does not differentiate 3ˣ or xˣ, whose exponent changes with x. For a positive differentiable inner function F near a, the chain rule gives [F(x)ʳ]′ at a equal to r F(a)ʳ⁻¹F′(a).

The exponent formula has a real domain

A reduced rational exponent p/q with odd positive q has a real negative-input branch: x^(p/q)=(−1)ᵖ|x|^(p/q) for x<0. The positive-power result and chain rule for −x give its derivative (p/q)x^(p/q−1) at every negative input. For even q there is no negative-input real root; arbitrary irrational powers are treated here on x>0. Reducing p/q first avoids assigning contradictory branches to the same exponent.

At zero, do not substitute into a formula containing a negative power. If r>0, extending the positive branch by F(0)=0 gives a right quotient hʳ⁻¹. For r>1 this approaches zero; for r=1 it is 1; for 0<r<1 it grows without a finite limit. To justify the first and last claims, the positive power hᵗ with t>0 tends to zero: for tolerance ε>0, 0<h<ε^(1/t) implies hᵗ<ε by positive-power order. Its reciprocal cannot approach a finite number. Nonpositive exponents exclude zero unless a separately defined constant extension is specified.

An odd-denominator rational branch with r>1, extended by value zero, has derivative zero from both sides at zero, since its quotient has absolute value |h|ʳ⁻¹. A positive-domain power extended to x≥0 supplies only a right-hand endpoint derivative. The diagrams distinguish an unavailable side from a side with no finite derivative; neither is silently recorded as zero.

A missing hypothesis asks for another argument

A right-hand derivative uses the same quotient and actual f(a), restricted to h>0; a left-hand derivative uses h<0. Require domain inputs arbitrarily close on the named side. On a shared one-sided interval the sum, product and guarded quotient proofs use the same identities. For a one-sided chain argument, all nearby inner values must lie in the outer domain, and any nonzero inner increments must approach on sides where the required finite outer derivative is known. Inner continuity on the tested side is still needed and follows from the finite inner derivative.

The hypotheses of the chain rule are sufficient, not necessary. With outer g(u)=|u| at zero and inner f(x)=x², the outer derivative fails, but g(f(x))=x² has derivative zero. Establish this by its own quotient h, rather than writing a nonexistent outer derivative times zero.

With outer √u and inner x², the composition is √(x²)=|x| on the real line. At zero its left quotient is −1 and right quotient 1, so no finite two-sided derivative exists. Restricting the input domain to x≥0 changes the composition to x and gives right derivative 1. The square root’s outer quotient at zero is unbounded, so the finite chain theorem does not apply in either case. Zero times an undefined derivative is not a valid calculation.

Name the operation, then check each layer

Choose a supplied formula and input in the explorer. Predict the finite derivative or the domain obstruction before revealing exact records and reasoning. The input formula, domain, base input and nearby test inputs are visible first; answers, output records and derivative markers appear after reveal.

For a product, keep both changing factors. For a quotient, check the original denominator at the input. For a composition, list the successive inputs and evaluate each derivative there. For a power, identify a fixed exponent and its real domain. A finite record table is an arithmetic check; the proved identities and error bounds establish the limiting statement.

Worked examples

Try it yourself

1. Differentiate 4x³−7x+2. What is the derivative at −1?

   Show the answer and reasoning

   12x²−7; at −1 it is 5.

   The coefficients 4, −7 and 2 are fixed. Apply integer powers, fixed scalars and the sum rule; the constant contributes zero.

2. A learner differentiates x·x² as x·2x=2x². Find the missing term.

   Show the answer and reasoning

   The derivative is 3x²; the missing term is x².

   Both factors depend on x. The product rule gives 1·x²+x·2x=3x², agreeing with the derivative of x³.

3. Find the derivative of (x+1)(2x+1) at x=−1.

   Show the answer and reasoning

   −1.

   The product rule gives 4x+3. At −1 the first factor is zero, but its derivative is 1: 1·(−1)+0·2=−1.

4. Differentiate x/(x+2), keeping its real domain.

   Show the answer and reasoning

   2/(x+2)² for x≠−2.

   The numerator is 1·(x+2)−x·1=2. Reversing the numerator order would change the sign. The original excluded input remains excluded.

5. Find the derivative of x^(−2) at x=2. Is x=0 allowed?

   Show the answer and reasoning

   −1/4; zero is excluded.

   The reciprocal of x² gives derivative −2x^(−3). At two this is −2/8. Neither the original function nor this derivative is defined at zero.

6. For f(x)=2x and g(u)=u², find the derivative of g(f(x)) at x=−1.

   Show the answer and reasoning

   −8.

   The outer input is f(−1)=−2, so g′ there is −4. Multiply by f′(−1)=2 to obtain −8; the expanded composite is 4x².

7. Differentiate √(3x+1) at x=1. State where the formula supplies a finite two-sided derivative.

   Show the answer and reasoning

   3/4 at one; the formula 3/[2√(3x+1)] applies for x>−1/3.

   At one the inner value is 4, where the outer derivative is 1/4. Multiply by 3. At the endpoint x=−1/3 the outer root is at zero and its right quotient is unbounded.

8. For a fixed irrational exponent r=√2, differentiate xʳ at x=1. Does the same expression define a real negative-input branch here?

   Show the answer and reasoning

   Derivative √2 at one; no negative-input branch is included.

   The proved fixed-real-exponent rule applies on x>0, giving r·1ʳ⁻¹=r. Arbitrary irrational powers are not assigned real negative-base values in this convention.

9. Extend x^(3/2) to x≥0 by F(0)=0. Which derivative exists at zero?

   Show the answer and reasoning

   The right-hand derivative is 0; a left-hand real-domain approach is unavailable.

   For h>0 the quotient is √h and tends to zero. There are no allowed h<0. This establishes a right endpoint derivative, not a two-sided derivative on an open interval.

10. Does the ordinary finite chain rule establish the derivative of |x²| at zero? Find that derivative.

    Show the answer and reasoning

    The theorem’s outer hypothesis fails, but the composite derivative is 0.

    The outer |u| has unequal side quotients at zero. However |x²|=x² for all real x, and its direct quotient at zero is h, which approaches zero.

11. Compare the derivative of √(x²) at zero on ℝ and on the restricted domain x≥0.

    Show the answer and reasoning

    No finite two-sided derivative on ℝ; right-hand derivative 1 on x≥0.

    The composition equals |x|. Its quotients are −1 on the left and 1 on the right. With only nonnegative inputs it equals x and the right quotient is 1. An undefined outer derivative cannot be multiplied by zero.

12. If f+g is differentiable at zero, must f and g each be differentiable there?

    Show the answer and reasoning

    No.

    Take f(x)=|x| and g(x)=−|x|. Each has unequal left and right quotients at zero, but f+g is constant zero. The sum rule is a sufficient forward theorem, not its converse.

What to carry forward

• A derivative uses the actual point value and a limit of nonzero input increments.
• Sum and fixed-scalar rules require finite derivatives at the same input; a variable factor needs the product rule.
• A product derivative is f′g+fg′. The corner term vanishes in the limit.
• A quotient derivative is (f′g−fg′)/g² where the original denominator at the input is nonzero.
• The chain rule is g′(f(a))f′(a); its remainder proof includes zero inner changes.
• The fixed-real-power rule r xʳ⁻¹ applies for x>0; negative branches and zero endpoints need their stated conditions.
• When a hypothesis fails, examine the actual quotient and domain instead of assigning a nonexistent derivative.