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The Derivative from First Principles · Tutorial 343 of 1495

Quadratic Derivatives by Factoring

Learn to derive quadratic derivatives at specified and arbitrary inputs by simplifying the difference quotient before taking its limit.

Intermediate 9 min read

What You'll Learn

  • Set up a derivative at an input using the increment form of the difference quotient.
  • Expand and factor quadratic difference quotients correctly.
  • Explain why the increment may be cancelled only when it is nonzero.
  • Derive the derivative of a quadratic polynomial at an arbitrary input.
  • Check specified-input and arbitrary-input calculations using the simplified quotient.

Why Factoring Helps

A derivative measures the limiting rate of change at an input. From first principles, that rate is defined using a limit of secant slopes. For a quadratic polynomial, direct substitution into the difference quotient initially produces a fraction with numerator and denominator both equal to zero. Factoring the numerator reveals why the quotient nevertheless has a finite limit.

The key is to keep two ideas separate: the increment \(h\) is nonzero in the difference quotient, but it approaches zero when we take the limit. We simplify the quotient for \(h\ne 0\) first; only then do we evaluate its limit as \(h\) approaches zero.

Definition: The derivative of a function \(f\) at an input \(x_0\), when the limit exists, is \(\displaystyle f'(x_0)=\lim_{h\to 0}\frac{f(x_0+h)-f(x_0)}{h}\). Here \(h\ne 0\) while the quotient is formed. The limit asks what values the quotient approaches as nonzero \(h\) gets arbitrarily close to zero.

The numerator compares the function value at \(x_0+h\) with its value at \(x_0\). Dividing by \(h\) gives the average rate of change over that input change. If the quotient has a finite limit, that limit is the derivative at \(x_0\).

A Specified Input: Expand, Subtract, Factor

Worked Example: A Quadratic at \(x=2\)

Find the derivative of \(f(x)=x^2+3x-2\) at \(x=2\) from the definition.

First calculate the function value at the specified input:

$$ f(2)=2^2+3(2)-2=4+6-2=8. $$

Now calculate the function value at \(2+h\), showing the expansion:

$$ \begin{aligned} f(2+h) &=(2+h)^2+3(2+h)-2\\ &=4+4h+h^2+6+3h-2\\ &=8+7h+h^2. \end{aligned} $$

Substitute these values into the difference quotient. For \(h\ne 0\), subtract and collect the terms in the numerator:

$$ \begin{aligned} \frac{f(2+h)-f(2)}{h} &=\frac{(8+7h+h^2)-8}{h}\\ &=\frac{7h+h^2}{h}\\ &=\frac{h(7+h)}{h}\\ &=7+h. \end{aligned} $$

Now take the limit. Since \(7+h\) approaches \(7\) as \(h\) approaches zero,

$$ f'(2)=\lim_{h\to 0}(7+h)=7. $$

Check: For every nonzero \(h\), the original quotient simplifies to \(7+h\). Its limit is indeed \(7\), so the derivative at \(2\) is \(7\).

Why the Increment Factor Cancels

In the example, the numerator became \(7h+h^2\), which factors as \(h(7+h)\). The denominator is also \(h\), so the factor can be cancelled provided \(h\ne 0\). That condition is part of the definition: the quotient is considered for nonzero increments and then its limit is taken as the increments approach zero.

Important: Cancelling \(h\) does not mean substituting \(h=0\) into the original quotient. The original quotient is undefined at \(h=0\). Cancellation gives an expression equal to that quotient for every \(h\ne 0\), and the limit of the simplified expression can then be evaluated.

This works because a limit depends on the values of an expression arbitrarily near the point, not on whether the expression is defined at the point itself. In this case, after cancellation the quotient is a simple linear expression in \(h\), whose limit can be read off directly.

Quadratics at Arbitrary Inputs

The same calculation works at any real input. Let \(f(x)=Ax^2+Bx+C\), where \(A\ne 0\), and let \(x_0\) be any real number. The polynomial is defined at \(x_0\) and at all nearby inputs. Expand the value at \(x_0+h\), subtract the value at \(x_0\), and factor the increment.

Theorem: If \(f(x)=Ax^2+Bx+C\), with \(A\ne 0\), then its derivative exists at every real input \(x_0\), and \(f'(x_0)=2Ax_0+B\).
1
Write the two function values.
\(f(x_0+h)=A(x_0+h)^2+B(x_0+h)+C\), while \(f(x_0)=Ax_0^2+Bx_0+C\).
2
Expand and subtract.
Use \((x_0+h)^2=x_0^2+2x_0h+h^2\), then subtract all of \(f(x_0)\).
3
Factor and cancel for \(h\ne 0\).
The numerator has a factor of \(h\), which cancels with the denominator.
4
Take the limit.
The remaining term that contains \(h\) approaches zero, leaving the derivative at \(x_0\).

Carrying out each step gives the complete derivation:

$$ \begin{aligned} f(x_0+h) &=A(x_0+h)^2+B(x_0+h)+C\\ &=A(x_0^2+2x_0h+h^2)+Bx_0+Bh+C\\ &=Ax_0^2+2Ax_0h+Ah^2+Bx_0+Bh+C, \end{aligned} $$
$$ \begin{aligned} \frac{f(x_0+h)-f(x_0)}{h} &=\frac{Ax_0^2+2Ax_0h+Ah^2+Bx_0+Bh+C -(Ax_0^2+Bx_0+C)}{h}\\ &=\frac{2Ax_0h+Ah^2+Bh}{h}\\ &=\frac{h(2Ax_0+Ah+B)}{h}\\ &=2Ax_0+Ah+B,\qquad h\ne 0. \end{aligned} $$

As \(h\to 0\), the term \(Ah\) approaches zero. Therefore,

$$ f'(x_0)=\lim_{h\to 0}(2Ax_0+Ah+B)=2Ax_0+B. $$

This proves the result for every real \(x_0\), not just for one chosen example. The constant term \(C\) disappears because the same \(C\) occurs in both function values and cancels in their difference.

Applying the Result Carefully

Worked Example: A Quadratic at \(x=-1\)

Find the derivative of \(g(x)=4x^2-5x+7\) at \(x=-1\) from the difference quotient.

First evaluate the function at \(-1\):

$$ g(-1)=4(-1)^2-5(-1)+7=4+5+7=16. $$

Next expand \(g(-1+h)\):

$$ \begin{aligned} g(-1+h) &=4(-1+h)^2-5(-1+h)+7\\ &=4(1-2h+h^2)+5-5h+7\\ &=4-8h+4h^2+5-5h+7\\ &=16-13h+4h^2. \end{aligned} $$

Substitute into the quotient, factor out \(h\), and cancel it for \(h\ne 0\):

$$ \begin{aligned} \frac{g(-1+h)-g(-1)}{h} &=\frac{(16-13h+4h^2)-16}{h}\\ &=\frac{-13h+4h^2}{h}\\ &=\frac{h(-13+4h)}{h}\\ &=-13+4h. \end{aligned} $$

Taking the limit gives

$$ g'(-1)=\lim_{h\to 0}(-13+4h)=-13. $$

Check: The simplified quotient is \(-13+4h\) for every nonzero \(h\), and its limit is \(-13\). Thus the derivative at \(-1\) is \(-13\).

Worked Example: A Quadratic at an Arbitrary Input

Find the derivative of \(q(x)=2x^2-3x+4\) at an arbitrary real input \(x\), using the difference quotient.

Compute the shifted value and subtract the original value:

$$ \begin{aligned} q(x+h) &=2(x+h)^2-3(x+h)+4\\ &=2(x^2+2xh+h^2)-3x-3h+4\\ &=2x^2+4xh+2h^2-3x-3h+4,\\[4pt] q(x+h)-q(x) &=(2x^2+4xh+2h^2-3x-3h+4) -(2x^2-3x+4)\\ &=4xh+2h^2-3h. \end{aligned} $$

For \(h\ne 0\), factor and cancel the increment:

$$ \begin{aligned} \frac{q(x+h)-q(x)}{h} &=\frac{4xh+2h^2-3h}{h}\\ &=\frac{h(4x+2h-3)}{h}\\ &=4x+2h-3. \end{aligned} $$

Now let \(h\) approach zero:

$$ q'(x)=\lim_{h\to 0}(4x+2h-3)=4x-3. $$

Check: The quotient simplifies to \(4x+2h-3\) for every nonzero \(h\), and its limit is \(4x-3\). This verifies the derivative at every real input \(x\).

A Common Mistake

A tempting but incorrect calculation is to substitute \(h=0\) into the original difference quotient and conclude that the derivative is undefined:

$$ \frac{f(x_0+0)-f(x_0)}{0}=\frac{0}{0}. $$

This only shows that the quotient itself is not defined at \(h=0\). The derivative is a limit of the quotient as nonzero \(h\) approaches zero; it is not the quotient evaluated at zero.

For example, for \(q(x)=2x^2-3x+4\), the quotient is \(\frac{h(4x+2h-3)}{h}\) when \(h\ne 0\). Cancel the common factor under that condition, then take the limit of \(4x+2h-3\). The limit is \(4x-3\). The cancellation does not make the original fraction defined at zero; it identifies a simpler expression with the same values for all nonzero increments.

Why This Method Matters

Factoring makes the defining limit manageable by exposing the part of the change that is proportional to the increment. Once that increment factor is cancelled, the remaining expression describes how the average rate of change behaves as the interval shrinks. Its limit gives the instantaneous rate of change, or equivalently the tangent slope at the input.

The method also makes clear what is special about quadratics: after expansion and subtraction, every surviving numerator term contains \(h\). Cancelling that factor leaves a term that approaches a fixed value as \(h\to 0\). For a quadratic, the calculation works at every real input, as the arbitrary-input derivation shows.

Check Your Understanding

For each question, pay attention to when the increment is nonzero and when the limit is taken.

  1. For \(f(x)=x^2-2x+5\), write and simplify the difference quotient at \(x_0=1\), then find \(f'(1)\).
  2. For \(g(x)=3x^2+4x-1\), calculate \(g(x+h)-g(x)\) and identify its factor of \(h\).
  3. Why is it valid to cancel \(h\) in the difference quotient, even though \(h\) approaches zero?
  4. Use the arbitrary-input result to find the derivative of \(p(x)=-2x^2+7x+3\) at \(x=4\).
  5. What does substituting \(h=0\) into the original difference quotient tell you, and what does it not tell you?

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