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Basic Differentiation Rules · Tutorial 447 of 1495

Polynomial Differentiation Workshop

Learn a reliable workflow for expanding and differentiating polynomials, evaluating derivatives, and checking the results with nearby secant slopes.

Beginner 10 min read

What You'll Learn

  • Expand products and combine like terms before differentiating.
  • Apply the power rule to each term of a polynomial.
  • Evaluate a symbolic derivative at a specified input.
  • Check derivative values using secant slopes from both sides.
  • Distinguish a numerical check from a proof of a derivative.

A Reliable Workflow for Polynomial Derivatives

A polynomial may be presented in a form that hides its individual powers. It might contain parentheses, products, or several terms that need to be collected. A dependable way to handle such expressions is to expand them first, combine like terms, and then differentiate each term.

This order keeps the calculation visible. It also makes it easier to evaluate the derivative at a point and to check whether the result is plausible using secant slopes. The power rule and the basic rules for sums and constant multiples are the tools; the main skill here is using them carefully in a mixed calculation.

Polynomial differentiation workflow: Expand the expression, combine like terms, differentiate each term, and only then substitute the requested input. Finally, check the value against secant slopes if a numerical check is useful.

The Term-by-Term Rule

A polynomial in \(x\) is a finite sum of constant multiples of nonnegative integer powers of \(x\). For example, \(3x^4-2x^2+5x-7\) is a polynomial. The power rule says that the derivative of \(a x^k\), where \(a\) is constant and \(k\) is a positive integer, is \(ka x^{k-1}\). A constant has derivative zero.

Rule: If \(p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\), then $$ p'(x)=na_nx^{n-1}+(n-1)a_{n-1}x^{n-2}+\cdots+2a_2x+a_1. $$ Here the coefficients \(a_0,\ldots,a_n\) are constants. The constant term \(a_0\) contributes zero, and terms with zero coefficients may simply be omitted.

In practice, each term is differentiated separately: multiply its coefficient by its exponent, then reduce the exponent by one. This rule applies to a polynomial once it has been written as a sum of powers. If the expression is given as a product, expand it first in this workshop; the product rule is a separate rule.

Why Differentiating Each Term Works

The reason the rule works is that the difference quotient of a finite sum is the sum of its difference quotients. The power rule supplies the limit for each individual power. Since a polynomial has only finitely many terms, the limit of the sum is the sum of those limits.

Derivation: Let \(p(x)=\sum_{k=0}^{n}a_kx^k\). For \(h\ne0\), the difference quotient at \(x\) is $$ \frac{p(x+h)-p(x)}{h} = \sum_{k=0}^{n}a_k\frac{(x+h)^k-x^k}{h}. $$ Taking the limit as \(h\to0\), the \(k=0\) term is zero, and for each \(k\ge1\), the power rule gives the limit \(k x^{k-1}\). Therefore $$ p'(x)=\sum_{k=1}^{n}k a_kx^{k-1}. $$ This proves the term-by-term formula for every polynomial.

This derivation uses the power rule for individual powers and the fact that limits distribute over finite sums. It does not require differentiating a product of polynomial factors. Expanding a product here is an algebra step that rewrites the expression as a sum before differentiation.

Worked Examples

Worked Example: Expand, Differentiate, and Check at \(x=1\)

Let \(f(x)=(x+2)(x^2-x+3)\). Find \(f'(1)\), then check the value with secant slopes on either side of \(1\).

First expand each part of the product:

$$ \begin{aligned} f(x) &=x(x^2-x+3)+2(x^2-x+3)\\ &=x^3-x^2+3x+2x^2-2x+6\\ &=x^3+x^2+x+6. \end{aligned} $$

Differentiate term by term. The derivative of \(x^3\) is \(3x^2\), the derivative of \(x^2\) is \(2x\), the derivative of \(x\) is \(1\), and the derivative of \(6\) is \(0\). Thus

$$ f'(x)=3x^2+2x+1. $$

Substitute \(x=1\):

$$ f'(1)=3(1)^2+2(1)+1=3+2+1=6. $$

Now check with secant slopes. The function value at \(1\) is \(f(1)=1+1+1+6=9\). For a step \(h\ne0\), the secant slope from \(1\) to \(1+h\) is \(\frac{f(1+h)-f(1)}{h}\). With \(h=0.1\),

$$ f(1.1)=1.331+1.21+1.1+6=9.641, \qquad \frac{9.641-9}{0.1}=6.41. $$

With \(h=-0.1\),

$$ f(0.9)=0.729+0.81+0.9+6=8.439, \qquad \frac{8.439-9}{-0.1}=5.61. $$

The two secant slopes, \(6.41\) and \(5.61\), lie on opposite sides of the derivative value \(6\). They are consistent with the derivative, although neither secant slope has to equal it.

Worked Example: Keep Track of Subtraction and Like Terms

Let \(g(x)=4(2x^2-3x+1)-(x-2)(x+3)\). Find \(g'(-2)\), and check it using nearby secant slopes.

Expand the first part and the product being subtracted:

$$ 4(2x^2-3x+1)=8x^2-12x+4, $$ $$ (x-2)(x+3)=x^2+x-6. $$

Subtract the entire second expression, changing the sign of each of its terms:

$$ \begin{aligned} g(x) &=8x^2-12x+4-(x^2+x-6)\\ &=8x^2-12x+4-x^2-x+6\\ &=7x^2-13x+10. \end{aligned} $$

Differentiate:

$$ g'(x)=14x-13. $$

At \(x=-2\),

$$ g'(-2)=14(-2)-13=-28-13=-41. $$

For a check, first calculate \(g(-2)\) from the original expression. The first part is \(4(2(4)-3(-2)+1)=4(8+6+1)=60\), while the product is \((-2-2)(-2+3)=(-4)(1)=-4\). Hence \(g(-2)=60-(-4)=64\).

At \(-1.9\), the expanded polynomial gives

$$ g(-1.9)=7(3.61)-13(-1.9)+10=25.27+24.7+10=59.97, $$ $$ \frac{g(-1.9)-g(-2)}{-1.9-(-2)} = \frac{59.97-64}{0.1} = -40.3. $$

At \(-2.1\),

$$ g(-2.1)=7(4.41)-13(-2.1)+10=30.87+27.3+10=68.17, $$ $$ \frac{g(-2.1)-g(-2)}{-2.1-(-2)} = \frac{68.17-64}{-0.1} = -41.7. $$

These secant slopes, \(-40.3\) and \(-41.7\), are close to the derivative value \(-41\). The sign in the second denominator matters: the point \(-2.1\) is \(0.1\) to the left of \(-2\).

Worked Example: Expand Powers Before Evaluating

Let \(q(x)=(x-1)^3+2(x+1)^2-3x+5\). Find \(q'(2)\), and use secant slopes at \(2\) to check it.

Expand the powers:

$$ (x-1)^3=x^3-3x^2+3x-1, \qquad 2(x+1)^2=2(x^2+2x+1)=2x^2+4x+2. $$

Now combine all terms, including the remaining \(-3x+5\):

$$ \begin{aligned} q(x) &=x^3-3x^2+3x-1+2x^2+4x+2-3x+5\\ &=x^3-x^2+4x+6. \end{aligned} $$

Differentiate each term and evaluate at \(2\):

$$ q'(x)=3x^2-2x+4, \qquad q'(2)=3(2)^2-2(2)+4=12-4+4=12. $$

For the secant checks, \(q(2)=8-4+8+6=18\). At \(2.1\),

$$ q(2.1)=9.261-4.41+8.4+6=19.251, \qquad \frac{19.251-18}{0.1}=12.51. $$

At \(1.9\),

$$ q(1.9)=6.859-3.61+7.6+6=16.849, \qquad \frac{16.849-18}{-0.1}=11.51. $$

The slopes \(12.51\) and \(11.51\) bracket \(12\), supporting the derivative calculation.

Secant Slopes Are Checks, Not Substitutes

For a function \(p\), a secant slope between inputs \(a\) and \(a+h\), with \(h\ne0\), is \(\frac{p(a+h)-p(a)}{h}\). It measures the average change in function value across that interval. The derivative \(p'(a)\), when it exists, is the limit of these slopes as \(h\) approaches zero.

$$ p'(a)=\lim_{h\to0}\frac{p(a+h)-p(a)}{h}. $$

For a polynomial, the derivative exists at every real input. Choosing small positive and negative values of \(h\) can therefore provide a useful numerical check. But any one secant slope uses two distinct inputs, so it is generally only an approximation to the derivative at \(a\). A close match is evidence that the calculation is plausible; it is not a replacement for the symbolic differentiation.

A Common Error: Losing Terms When Subtracting

In the second example, the expression \(-(x^2+x-6)\) must become \(-x^2-x+6\). A common incorrect line is \(-(x^2+x-6)=-x^2+x-6\), which changes only the first sign. That would give the wrong polynomial and consequently the wrong derivative.

The correction is to distribute the negative sign to every term inside the parentheses:

$$ -(x^2+x-6)=-x^2-x+6. $$

Then combine like terms only after the signs are correct. It is also important not to confuse a derivative value with a function value: in general, \(p'(a)\) and \(p(a)\) are different quantities. Write the derivative first, then substitute \(a\) into that derivative.

Summary

For mixed polynomial differentiation, first turn the expression into a sum of powers by expanding and collecting like terms. Apply the power rule to each term, remember that constants differentiate to zero, and substitute the requested input into the derivative. Secant slopes from nearby points can check the value, provided their denominators retain the correct signs. The derivative is the limiting slope, not usually any single nearby secant slope.

Check Your Understanding

Work through these questions, keeping expansion, differentiation, and evaluation as separate steps.

  1. Expand \(r(x)=(x+1)(x^2+2x+4)\), then find and simplify \(r'(x)\).
  2. For \(s(x)=3(2x^2-x)-(x+2)(x-1)\), write the expanded polynomial and find \(s'(1)\).
  3. Differentiate \(t(x)=x^4-3x^2+2x-8\). Which term contributes zero, and why?
  4. If \(p'(a)=5\), must the secant slope from \(a\) to \(a+0.1\) equal \(5\)? Explain the distinction.
  5. For \(u(x)=x^2+2x\), calculate the secant slope from \(x=1\) to \(x=1.1\), then compare it with \(u'(1)\).

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