Why Linearity Matters
A derivative measures the limiting change in a function near a point. When two functions are added, or when a function is multiplied by a fixed number, it is natural to ask how their derivatives are related. The answer is that differentiation respects these operations: under the right conditions, the derivative of a sum is the sum of the derivatives, and a fixed constant can be taken outside the derivative.
These facts are often used as rules for calculation. Here we derive them directly from the difference quotient. The derivation shows both why the rules work and what assumptions they require. We will then check them using polynomial calculations at particular points.
The Difference Quotient at a Point
Let \(f\) be defined on an open interval containing a point \(a\). For a nonzero increment \(h\) sufficiently close to zero, the difference quotient of \(f\) at \(a\) is
If the limit of this quotient exists as a finite real number when \(h\to 0\), then \(f\) is differentiable at \(a\), and that limit is the derivative of \(f\) at \(a\):
The requirement \(h\ne 0\) applies to the quotient itself. The limit examines its values for nonzero increments approaching zero; it does not require evaluating the quotient at \(h=0\).
For the sum rule, both functions must be defined in a common open interval around \(a\), and each must be differentiable at \(a\). The common interval ensures that the values of both functions, and hence their sum, are defined for the small increments used in the quotient.
Deriving the Sum Rule
Suppose \(f\) and \(g\) are differentiable at \(a\). The difference quotient for their sum can be separated algebraically because both differences have the same denominator:
By differentiability, each quotient on the right has a finite limit as \(h\to 0\). The limit law for sums says that the limit of a sum of two functions is the sum of their limits when both finite limits exist. Applying that law gives the derivative of the sum.
The existence conditions matter: the limit law is being applied to two quotients that each have a finite limit. The algebraic separation alone does not establish that either limit exists.
Deriving the Constant-Multiple Rule
Let \(c\) be a fixed real number, and suppose \(f\) is differentiable at \(a\). Since \((cf)(x)=c f(x)\), the difference quotient is
Taking the limit, the constant-multiple law for finite limits allows the fixed number \(c\) to remain outside the limit. The resulting limit exists because \(f\) is differentiable at \(a\).
The word constant is essential. The proof takes the same number \(c\) out of the quotient for every increment \(h\). A coefficient that changes with the input is not covered by this argument.
Combining the Rules
Repeated use of the sum rule and the constant-multiple rule gives a useful form of linearity. If \(f\) and \(g\) are differentiable at \(a\), and \(c\) and \(d\) are fixed real numbers, then
Indeed, the constant-multiple rule gives \((cf)'(a)=c f'(a)\) and \((dg)'(a)=d g'(a)\). Both \(cf\) and \(dg\) are differentiable at \(a\), so the sum rule applies to them:
This conclusion concerns sums and fixed scalar multiples. The proof does not establish a rule for multiplying two variable functions together.
Worked Examples from Difference Quotients
Worked Example: A Sum of Two Quadratic Polynomials
Let \(p(x)=x^2+1\) and \(q(x)=3x^2-2x\). Compute the derivative of \(p+q\) at \(a=2\) directly from its difference quotient, and compare it with the sum of the separate derivatives.
First evaluate the functions at \(2\):
Thus \((p+q)(2)=5+8=13\). For a nearby input \(2+h\), expand each polynomial:
Adding these values gives \((p+q)(2+h)=13+14h+4h^2\). Therefore, for \(h\ne 0\),
Taking \(h\to 0\) gives \((p+q)'(2)=14\). To check against the separate quotients, for \(p\) we have
For \(q\), we have
The separate limits add to \(4+10=14\), agreeing with the direct calculation. Substitution into the sum rule is therefore verified at this point.
Worked Example: A Negative Constant Multiple
Let \(f(x)=x^3-2x\). Find the derivative of \(-2f\) at \(a=1\) by computing the quotient directly.
First, \(f(1)=1^3-2(1)=-1\), so \((-2f)(1)=2\). Expand \(f(1+h)\):
It follows that \((-2f)(1+h)=2-2h-6h^2-2h^3\). The difference quotient is
Taking \(h\to 0\) gives \((-2f)'(1)=-2\). For a direct check of the constant-multiple rule, the quotient for \(f\) itself is
Thus \(f'(1)=1\), and multiplying by the fixed constant gives \(-2f'(1)=-2(1)=-2\), as found directly.
Worked Example: A Linear Combination at a Negative Point
Let \(p(x)=x^2\), \(q(x)=x^3\), and \(r(x)=2p(x)-3q(x)\). Compute \(r'(-1)\) directly and then check the result using the linearity rules.
At \(x=-1\), \(p(-1)=1\) and \(q(-1)=-1\), so \(r(-1)=2(1)-3(-1)=5\). At \(-1+h\), the expansions are
Consequently,
For \(h\ne 0\), the quotient simplifies to
Its limit is \(r'(-1)=-13\). To check using the separate quotients,
Linearity predicts \(r'(-1)=2(-2)-3(3)=-4-9=-13\), which agrees with the direct quotient limit.
A Common Mistake: Treating a Variable Coefficient as Constant
The constant-multiple rule allows a fixed number \(c\) to be taken outside a difference quotient. It does not allow a coefficient that depends on the input to be treated as fixed. For instance, an incorrect argument for \(x^2\) might say, “This is \(x\) times \(x\), so take one \(x\) outside and the derivative is just \(x\).” At \(a=2\), this would give \(2\).
The direct quotient shows the error:
The correct derivative at \(2\) is \(4\), not \(2\). The coefficient in a constant-multiple rule must be the same fixed number for every input and every small increment \(h\). Here the proposed coefficient is the variable \(x\), so the constant-multiple proof does not apply.
When the Linearity Rules Apply
Before using linearity at a point, check the hypotheses. For the sum rule, both functions must be differentiable at that point. For the constant-multiple rule, the function must be differentiable there and the multiplier must be a fixed real number. These conditions guarantee the finite limits needed in the derivations.
The proofs work because difference quotients preserve addition and multiplication by a fixed number, and because finite limits obey the corresponding algebraic laws. In practical calculations, linearity can reduce a complicated polynomial quotient to simpler pieces. In a first-principles argument, the same calculations explain exactly why that reduction is valid.
Check Your Understanding
Use the difference quotient and the conditions stated in this tutorial to answer these questions.
- State the difference-quotient definition of \(f'(a)\), including what is required of its limit.
- In the sum-rule derivation, why must both individual difference quotients have finite limits?
- If \(g\) is differentiable at \(a\) and \(c\) is a fixed real number, what is \((cg)'(a)\)?
- For \(p(x)=x^2\) and \(q(x)=x\), compute the difference quotient of \(p+q\) at \(a=1\) and find its limit.
- Explain why the constant-multiple rule cannot be applied by treating \(x\) as a constant coefficient in \(x^2=x\cdot x\).
