A Polynomial Derivative Workshop from First Principles

From the Definition to a Calculation

A polynomial derivative problem is a useful place to practise the central idea of differentiation from first principles: compare the function value at an input with its value a small distance away, divide by that distance, and take a limit. The algebra can look different from one polynomial to another. Sometimes expanding powers makes the cancellation visible; sometimes factoring the numerator is shorter.

For a function \(f:\mathbb{R}\to\mathbb{R}\), the derivative at a real input \(a\), when it exists, is defined by the limit of its difference quotient.

The subtraction \(f(a+h)-f(a)\) is essential: it removes the constant part of the numerator, leaving terms that can be divided by \(h\). For a polynomial, expansion or factorization will expose a factor of \(h\), or an equivalent cancellation, so that the quotient has a finite limit.

A Practical Choice: Expand or Factor?

Use expansion when substituting \(a+h\) into the polynomial gives manageable powers and you want to collect terms by powers of \(h\). Use factorization when the numerator has a recognizable difference of powers or another simple factorization. In either approach, keep the difference quotient intact until the cancellation is justified.

Worked Example: A Cubic at a Specified Input

Worked Example: Factoring a Difference of Powers

Worked Example: A Cubic at an Arbitrary Base Point

Why the Method Works for Every Polynomial

The examples reflect a general fact: every polynomial has a derivative at every real input, and its difference quotient can be simplified to a polynomial in \(h\). The binomial theorem gives a direct justification. For a nonnegative integer \(k\), it states that \((a+h)^k=\sum_{j=0}^{k}\binom{k}{j}a^{k-j}h^j\). In the difference \((a+h)^k-a^k\), the \(j=0\) term cancels, so every remaining term contains \(h\).

To prove the statement, apply the binomial theorem to each term of \(p(a+h)\), then subtract \(p(a)\):

The constant term, \(k=0\), contributes zero. In every other summand, each term has a factor \(h\). Since \(h\ne0\) in the difference quotient, divide each term by \(h\):

For each fixed \(a\), this is a finite sum. When \(h\to0\), the terms with \(j=1\) remain, while terms with \(j\ge2\) contain a positive power of \(h\) and tend to zero. Since \(\binom{k}{1}=k\), the limit is \(\sum_{k=1}^{n}k c_k a^{k-1}\). This proves the formula and, because the limit exists for every real \(a\), proves differentiability everywhere for real polynomials.

A Common Mistake: Forgetting to Subtract the Function Value

A frequent error is to divide \(f(a+h)\) by \(h\) instead of forming \(f(a+h)-f(a)\). For example, let \(r(x)=x^2\) and \(a=3\). The incorrect calculation is

This is not the difference quotient: it omits the required subtraction of \(r(3)=9\). The term \(9/h\) does not tend to zero as \(h\to0\), so discarding it to claim the answer is \(6\) is invalid. The correct calculation is

The correction follows the definition exactly: subtract the value at the base point first, simplify the whole numerator, and only then divide by \(h\).

What to Check in Your Own Work

Every polynomial derivative calculation from first principles should make three points visible: the correct base-point value, the cancellation that produces a quotient defined for \(h\ne0\), and the limit of the simplified expression. Expansion is often more informative when the base point is arbitrary; factorization can be more efficient when the numerator is a familiar difference. Both approaches are valid only when they preserve the original difference quotient.

The general polynomial result is useful because it explains why the workshop method always succeeds: after subtraction, the numerator has no constant term in \(h\). Dividing by \(h\) leaves a finite expression whose limit can be evaluated. The proof does not replace the definition; it shows why the definition gives a derivative for every polynomial at every real input.