Sum Rule

Adding Functions Before Differentiating

The Derivative of a Polynomial theorem differentiates a polynomial by handling its finitely many terms. The reason this works is that the difference quotient of a finite sum is itself the sum of the corresponding difference quotients. The Sum Rule makes this reasoning available for arbitrary functions that are differentiable at the point in question. Its hypotheses are local: the two functions need only be differentiable at the same point, not throughout their entire domains.

Let $I$ be an interval and let $a$ be an interior point of $I$. If $f$ and $g$ are defined on $I$, their sum is the function $f+g$ defined by $(f+g)(x)=f(x)+g(x)$. To find its derivative at $a$, we compare its value at $a+h$ with its value at $a$. The key algebraic step is to separate the change in $f$ from the change in $g$.

Theorem (Sum Rule): Let $I$ be an interval, let $a$ be an interior point of $I$, and let $f,g:I\to\mathbb{R}$ be differentiable at $a$. Then $f+g$ is differentiable at $a$, and $$ (f+g)'(a)=f'(a)+g'(a). $$

Proof from the Difference Quotient

Proof. For every nonzero $h$ small enough that $a+h\in I$, the difference quotient of the sum satisfies the exact identity

\frac{(f+g)(a+h)-(f+g)(a)}{h} = \frac{f(a+h)-f(a)}{h} + \frac{g(a+h)-g(a)}{h}.

Indeed, $(f+g)(a+h)=f(a+h)+g(a+h)$ and $(f+g)(a)=f(a)+g(a)$, so subtracting and dividing by $h$ gives the displayed equality. Since $f$ is differentiable at $a$, the first quotient tends to $f'(a)$ as $h\to0$. Since $g$ is differentiable at $a$, the second tends to $g'(a)$. The sum law for limits therefore gives

\lim_{h\to0} \frac{(f+g)(a+h)-(f+g)(a)}{h} = f'(a)+g'(a).

This limit exists and is finite, so $f+g$ is differentiable at $a$, with derivative $f'(a)+g'(a)$. $\square$

The proof uses the derivative definition and the sum law for limits. Differentiability of each summand supplies the two limits; the exact algebraic identity lets us combine them. The conclusion concerns the derivative at $a$. If $f$ and $g$ are differentiable at every point of $I$, then the same argument applies at each interior point, giving the function identity $(f+g)'=f'+g'$ there.

Worked Example: Adding Two Polynomials

Let

f(x)=x^4-3x,\qquad g(x)=2x^3+x^2-6.

Both functions are polynomials, so the Derivative of a Polynomial theorem gives

f'(x)=4x^3-3,\qquad g'(x)=6x^2+2x.

The Sum Rule applies at every real number. At $a=-1$, the separate derivatives are

$$ f'(-1)=4(-1)^3-3=-4-3=-7, $$

$$ g'(-1)=6(-1)^2+2(-1)=6-2=4. $$

Thus

$$ (f+g)'(-1)=f'(-1)+g'(-1)=-7+4=-3. $$

As a check, $f(x)+g(x)=x^4+2x^3+x^2-3x-6$, whose polynomial derivative is $4x^3+6x^2+2x-3$. At $-1$, this equals $-4+6-2-3=-3$, in agreement with the Sum Rule.

Worked Example: A Polynomial and a Reciprocal Function

Consider the functions $f(x)=x^2$ and $g(x)=1/x$ on the interval $I=(0,\infty)$. The point $a=2$ lies in the interior of this interval. By the Power Rule and the Power Rule for Negative Integer Exponents,

f'(x)=2x,\qquad g'(x)=-x^{-2}=-\frac{1}{x^2}

for $x>0$. The Sum Rule yields

(f+g)'(2)=f'(2)+g'(2)=4-\frac{1}{4}=\frac{15}{4}.

The domain matters: $g$ is not defined at zero, so this application is made on an interval that excludes zero. At $2$, both functions are differentiable and the sum is defined, which are the relevant conditions here.

Finite Sums

The two-function rule can be applied repeatedly. This gives a finite-sum version, which explains the term-by-term calculation for polynomials without requiring a separate argument for each number of terms.

Theorem (Finite Sum Rule): Let $I$ be an interval, let $a$ be an interior point of $I$, and let $f_1,\ldots,f_n:I\to\mathbb{R}$ be differentiable at $a$, where $n\geq1$ is an integer. Then $F=\sum_{j=1}^{n}f_j$ is differentiable at $a$, and $$ F'(a)=\sum_{j=1}^{n}f_j'(a). $$

Proof. We use induction on the positive integer $n$. For $n=1$, $F=f_1$, so the conclusion is exactly the assumed differentiability of $f_1$. Now suppose the statement holds for a sum of $n$ differentiable functions. For $n+1$ functions, define $G=\sum_{j=1}^{n}f_j$. By the induction hypothesis, $G$ is differentiable at $a$ and $G'(a)=\sum_{j=1}^{n}f_j'(a)$. The Sum Rule applied to $G$ and $f_{n+1}$ shows that $F=G+f_{n+1}$ is differentiable at $a$, with

F'(a)=G'(a)+f_{n+1}'(a) =\sum_{j=1}^{n}f_j'(a)+f_{n+1}'(a) =\sum_{j=1}^{n+1}f_j'(a).

This proves the statement for $n+1$, completing the induction. $\square$

Worked Example: A Sum with Several Terms

Let

$$ P(x)=3x^5-4x^3+2x^2-7x+9. $$

Apply the Finite Sum Rule to the five terms. The constant term has derivative zero, and the polynomial derivative formula gives

$$ P'(x)=15x^4-12x^2+4x-7. $$

At $x=1$, the derivative is

$$ P'(1)=15-12+4-7=0. $$

The same value results by differentiating each term and adding their derivatives at $1$: $15+(-12)+4+(-7)+0=0$. The finite-sum theorem justifies this procedure because there are finitely many differentiable terms.

Checking Differentiability Directly at a Point

The Sum Rule applies even when a formula is defined in pieces, provided each summand is differentiable at the point being considered. In such cases, it can be useful to check the derivative from its definition rather than rely on a derivative formula for the entire function.

Worked Example: Two Functions Differentiable at Zero

Define $f:\mathbb{R}\to\mathbb{R}$ by $f(x)=x^2$ for $x\geq0$ and $f(x)=-x^2$ for $x<0$, and let $g(x)=x^3$. Since $f(0)=g(0)=0$, for nonzero $h$ the difference quotients at zero are

\frac{f(h)-f(0)}{h} = \begin{cases} h,&h>0,\\ -h,&h<0, \end{cases} \qquad \frac{g(h)-g(0)}{h}=h^2.

Both expressions tend to zero as $h\to0$, so $f'(0)=0$ and $g'(0)=0$. The Sum Rule therefore gives $(f+g)'(0)=0$. Directly, for $h>0$ the sum's difference quotient is $h+h^2$, and for $h<0$ it is $-h+h^2$. Each tends to zero, confirming the conclusion from the definition.

What the Hypotheses Do—and Do Not—Say

The Sum Rule is a sufficient condition: if both functions are differentiable at $a$, then their sum is differentiable there. It does not say that differentiability of the sum forces differentiability of each summand. Cancellation can make a sum differentiable even when its separate terms are not.

Worked Example: A Differentiable Sum of Nondifferentiable Terms

Define $u(x)=|x|$ and $v(x)=-|x|$ on $\mathbb{R}$. At zero, the difference quotient for $u$ is

\frac{u(h)-u(0)}{h}=\frac{|h|}{h} = \begin{cases} 1,&h>0,\\ -1,&h<0. \end{cases}

The one-sided values do not approach the same limit, so $u$ is not differentiable at zero. The quotient for $v$ is $-|h|/h$, whose one-sided values are $-1$ and $1$, so $v$ is not differentiable there either. Nevertheless, $u(x)+v(x)=0$ for every $x$, and the zero function is differentiable at zero with derivative zero. Thus the Sum Rule cannot be reversed.

A different common mistake is to apply the rule when a summand fails to be differentiable at the point. The theorem's conclusion is guaranteed only when both derivatives exist there. The last example shows why it is important to read the implication in the correct direction: the hypotheses are enough to establish differentiability of the sum, but they are not necessary in every individual case.

The algebra behind the rule is straightforward, but its logical structure is worth retaining. First establish that both difference-quotient limits exist. Then split the quotient of the sum into those two quotients, and use the sum law for limits. This pattern is the basis for differentiating finite sums and for the broader rules that follow.

Check Your Understanding

Use the Sum Rule and the derivative definition where appropriate to answer these questions.

State the hypotheses needed to conclude that $f+g$ is differentiable at an interior point $a$.
If $f'(a)=-3$ and $g'(a)=8$, what is $(f+g)'(a)$?
For $F(x)=x^4-2x^2+5x$, find $F'(x)$ by treating $F$ as a finite sum.
Why does the Sum Rule proof require the sum law for limits?
Can a sum be differentiable at a point if one or both summands are not? Give an example and explain why this does not contradict the theorem.

What You'll Learn