Derivative of a Polynomial

From Powers to Polynomials

The Power Rule determines the derivative of each integer power $x^k$. A polynomial is a finite combination of such powers, so it is natural to expect that its derivative can be found by differentiating each term. Here we establish that formula directly from the difference quotient: the finite number of terms lets us take their limits one at a time. This gives a result for polynomials without assuming a general Sum Rule, which will be developed next.

Definition: A real polynomial is a function $p:\mathbb{R}\to\mathbb{R}$ of the form $p(x)=c_0+c_1x+\cdots+c_nx^n=\sum_{k=0}^{n}c_kx^k$, where $n$ is a nonnegative integer and $c_0,\ldots,c_n$ are real coefficients. The zero polynomial is the function that is identically zero. For a nonzero polynomial, its degree is the largest index $n$ for which $c_n\ne0$; its leading coefficient is $c_n$.

Some coefficients in a displayed polynomial may be zero, so a term such as $x^4$ need not be followed by nonzero $x^3,x^2,$ or $x$ terms. The degree is determined by the highest power with a nonzero coefficient, not by the number of terms that appear. The zero polynomial has no degree under this definition, because it has no nonzero coefficient.

The Derivative Formula

Theorem (Derivative of a Polynomial): Let $p(x)=\sum_{k=0}^{n}c_kx^k$ be a real polynomial. Then $p$ is differentiable at every $a\in\mathbb{R}$, and $$ p'(a)=\sum_{k=1}^{n}k c_k a^{k-1}. $$ If $n=0$, the sum on the right is empty and is understood to equal $0$. Equivalently, the derivative function is the polynomial $p'(x)=\sum_{k=1}^{n}k c_kx^{k-1}$.

Proof. Fix $a\in\mathbb{R}$. For every nonzero $h$, the difference quotient for $p$ can be written as the finite sum

\frac{p(a+h)-p(a)}{h} = \sum_{k=0}^{n}c_k\frac{(a+h)^k-a^k}{h}.

For $k=0$, the numerator is $1-1=0$, so this term is zero for every nonzero $h$. For each $k\geq1$, the Power Rule for Nonnegative Integer Exponents shows that

\lim_{h\to0}\frac{(a+h)^k-a^k}{h}=k a^{k-1}.

Multiplication by the fixed coefficient $c_k$ gives a limit of $k c_k a^{k-1}$. A finite sum of functions with limits has limit equal to the sum of those limits. Applying this limit law to the difference quotient above, including the zero term for $k=0$, yields

\lim_{h\to0}\frac{p(a+h)-p(a)}{h} = \sum_{k=1}^{n}k c_k a^{k-1}.

This limit exists for every real $a$, so $p$ is differentiable everywhere and its derivative has the stated value. Since the formula holds at every $a$, it also gives the derivative function $p'(x)=\sum_{k=1}^{n}k c_kx^{k-1}$. When $n=0$, the original function is constant and the formula gives the empty sum, or $0$, in agreement with the Derivative of a Constant Function Theorem. $\square$

The proof uses only a finite sum. It does not assume that an infinite sum can be differentiated term by term, nor does it establish a rule for arbitrary functions. Its key steps are that each monomial has a known derivative by the Power Rule and that a finite sum preserves the limit of its terms.

Worked Example: A Polynomial with Missing Powers

Consider

$$ p(x)=5x^6-2x^3+7. $$

The coefficients of $x^5,x^4,x^2,$ and $x$ are zero. Applying the polynomial formula to the nonzero terms gives

p'(x)=6\cdot5x^5+3\cdot(-2)x^2=30x^5-6x^2.

In particular, at $x=-1$,

$$ p'(-1)=30(-1)^5-6(-1)^2=-30-6=-36. $$

The constant $7$ contributes zero to the derivative. The missing powers contribute no terms either, because their coefficients are zero.

Worked Example: Evaluating a Polynomial Derivative at Zero

Let

$$ q(x)=-4x^5+3x^2-8x+6. $$

The derivative formula gives

$$ q'(x)=-20x^4+6x-8. $$

At zero, the terms containing positive powers of $x$ vanish, leaving

q'(0)=-20\cdot0^4+6\cdot0-8=-8.

The constant term $6$ has derivative zero, but the linear term $-8x$ has derivative $-8$. This distinction is easy to miss when evaluating at zero: the value of the polynomial at zero is $q(0)=6$, while its derivative there is $q'(0)=-8$.

How the Degree Changes

Theorem (Degree of the Derivative): Let $p$ be a nonzero polynomial of degree $n$. If $n\geq1$, then $p'$ is a nonzero polynomial of degree $n-1$. If $n=0$, then $p'=0$.

Proof. Write $p(x)=\sum_{k=0}^{n}c_kx^k$, where $c_n\ne0$. If $n\geq1$, the derivative formula gives

p'(x)=\sum_{k=1}^{n}k c_kx^{k-1}.

The largest possible power in this sum is $x^{n-1}$, arising from the term with $k=n$. Its coefficient is $n c_n$. Since $n$ is a positive integer and $c_n\ne0$, $n c_n\ne0$. All other terms have powers at most $n-2$, so none can cancel the $x^{n-1}$ term. Thus $p'$ is nonzero and has degree exactly $n-1$. If $n=0$, then $p$ is a nonzero constant, and the derivative formula gives $p'=0$. $\square$

The conclusion concerns nonzero polynomials. It would be inaccurate to say that the derivative of every polynomial has degree one less: a nonzero constant has derivative zero, and the zero polynomial has no degree under our definition. For a polynomial of positive degree, however, the leading term guarantees the degree drop, even if many lower coefficients are zero.

Worked Example: Predicting the Derivative's Degree

Let

$$ r(x)=-3x^7+4x^2-11. $$

The leading coefficient is $-3\ne0$, so $r$ has degree $7$. The Degree of the Derivative Theorem predicts that $r'$ has degree $6$. The formula confirms this:

$$ r'(x)=7(-3)x^6+2(4)x=-21x^6+8x. $$

The coefficient of $x^6$ is $-21$, which is nonzero, so the derivative really does have degree $6$. The absent $x^5,x^4,x^3,$ and constant terms in the derivative do not affect its degree.

Applying the Formula with Care

The formula is easiest to use when each coefficient and exponent is handled separately: multiply the coefficient by its exponent, then reduce that exponent by one. The constant term corresponds to exponent zero and contributes nothing. For instance, a term $c_0x^0=c_0$ has derivative zero; it should not be treated as producing a term involving $x^{-1}$. The $k=0$ term is deliberately excluded from the derivative sum.

Polynomial term	Contribution to the derivative	Reason
$c_0$	$0$	It is constant.
$c_kx^k$, $k\geq1$	$k c_kx^{k-1}$	Apply the Power Rule and multiply by $c_k$.
A missing power $x^k$	$0$	Its coefficient is zero.

Every real polynomial is defined on all of $\mathbb{R}$, and the theorem shows that its derivative is also defined there. This follows because each nonnegative integer power has that domain and the difference-quotient limit exists at every real input. No input needs to be excluded, including zero or negative values.

A common pitfall is to remember the exponent change but overlook the coefficient: the derivative of $c_kx^k$ is $k c_kx^{k-1}$, not merely $c_kx^{k-1}$. Another is to infer the derivative's degree from the number of terms rather than the highest nonzero power. The theorem handles both issues by recording every coefficient and identifying the leading term.

The polynomial result is a useful bridge between the Power Rule and broader differentiation rules. Its proof already shows why a finite sum can be handled term by term: the difference quotient is a finite sum, and its limit is the sum of the individual limits. The next step is to formulate that reasoning as a general rule for sums of differentiable functions.

Check Your Understanding

Use the polynomial derivative formula and the degree result to answer the following questions.

Find the derivative of $4x^8-3x^4+2x-9$, and identify its degree.
Why does the derivative formula omit the $k=0$ term, and what would go wrong if it were written as a term involving $x^{-1}$?
If a nonzero polynomial has degree $5$, what is the degree of its derivative, and why can its leading derivative coefficient not vanish?
What is the derivative of a nonzero constant polynomial? Does it have a degree under the definition used here?
Explain which fact about limits permits the proof to pass from the monomial difference quotients to the polynomial difference quotient.

Polynomial term	Contribution to the derivative	Reason
\(c_0\)	\(0\)	It is constant.
\(c_kx^k\), \(k\geq1\)	\(k c_kx^{k-1}\)	Apply the Power Rule and multiply by \(c_k\).
A missing power \(x^k\)	\(0\)	Its coefficient is zero.

What You'll Learn