Finding a Quadratic Derivative at a Point

From Secant Slopes to a Derivative at One Input

A quadratic function curves, so its slope is not the same at every input. To find its instantaneous rate of change at one chosen input, we can examine the slopes of secant lines through that point and a nearby point. As the nearby point approaches the chosen input, those secant slopes approach the derivative.

This tutorial uses that limit directly. The main work is algebra: substitute a nearby input, expand the quadratic, and simplify the difference quotient. We will not use derivative rules. In particular, the calculation is made at a specified input, rather than finding a formula for the derivative function.

The Derivative at a Specified Input

Definition: The derivative of a function $f$ at the input $a$, when the limit exists, is the limit of the difference quotient: $$ f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}. $$ Here $h$ is a nonzero increment while the quotient is formed. We take the limit as $h$ approaches zero; we do not substitute $h=0$ into the quotient.

The numerator $f(a+h)-f(a)$ is the change in the function’s output between inputs $a$ and $a+h$. Dividing by the input change $h$ gives the slope of the secant line between those two points. Taking the limit lets the second point approach the first, producing the tangent slope when that limit exists.

For a quadratic, the expression $f(a+h)$ contains a squared binomial. Expanding it exposes a constant term equal to $f(a)$, along with terms involving $h$. Subtracting $f(a)$ cancels that constant term. The remaining terms contain a factor of $h$, which can be canceled against the denominator before taking the limit.

A General Justification for Quadratics

We can see why the method works for any quadratic by considering $q(x)=Ax^2+Bx+C$, where $A$, $B$, and $C$ are fixed real numbers. Let $r$ be one fixed input. We will calculate the derivative at $r$ from the definition, not by applying a derivative rule.

Write the difference quotient.
Use $q(r+h)$ for the nearby output and subtract $q(r)$.

Expand and subtract.
The constant terms cancel, leaving terms with a factor of $h$.

Cancel and take the limit.
For $h\ne 0$, cancel the common factor of $h$, then let $h$ approach zero.

Here is the complete calculation. The input $r$ stays fixed throughout:

\begin{aligned} q'(r) &=\lim_{h\to 0}\frac{q(r+h)-q(r)}{h}\\ &=\lim_{h\to 0} \frac{\bigl(A(r+h)^2+B(r+h)+C\bigr)-\bigl(Ar^2+Br+C\bigr)}{h}\\ &=\lim_{h\to 0} \frac{A(r^2+2rh+h^2)+Br+Bh+C-Ar^2-Br-C}{h}\\ &=\lim_{h\to 0}\frac{2Arh+Ah^2+Bh}{h}\\ &=\lim_{h\to 0}(2Ar+Ah+B)\\ &=2Ar+B. \end{aligned}

The cancellation in the fourth line is essential: the terms $Ar^2$, $Br$, and $C$ appear once with a positive sign and once with a negative sign. In the next line, factoring out $h$ from the numerator allows cancellation because the difference quotient is considered for $h\ne 0$. The resulting expression $2Ar+Ah+B$ has a limit as $h\to 0$, namely $2Ar+B$.

Result: For $q(x)=Ax^2+Bx+C$, the derivative at a specified input $r$, found from the limit definition, is $q'(r)=2Ar+B$. This calculation relies on expanding the quadratic and canceling before taking the limit; it does not permit substituting $h=0$ into the original quotient.

Worked Examples

Worked Example: A Quadratic with Positive and Negative Terms

Find the derivative of $f(x)=2x^2-3x+1$ at $x=2$ using the limit definition.

First evaluate the function at the specified input:

$$ f(2)=2(2)^2-3(2)+1=8-6+1=3. $$

Now form the difference quotient, keeping $h\ne 0$ while simplifying:

\begin{aligned} f'(2) &=\lim_{h\to 0}\frac{f(2+h)-f(2)}{h}\\ &=\lim_{h\to 0} \frac{\bigl(2(2+h)^2-3(2+h)+1\bigr)-3}{h}\\ &=\lim_{h\to 0} \frac{2(4+4h+h^2)-6-3h+1-3}{h}\\ &=\lim_{h\to 0}\frac{8+8h+2h^2-6-3h+1-3}{h}\\ &=\lim_{h\to 0}\frac{5h+2h^2}{h}\\ &=\lim_{h\to 0}(5+2h)\\ &=5. \end{aligned}

Check the result using the simplified quotient: as $h$ approaches zero, $5+2h$ approaches $5$. Thus the difference quotient has limit $5$, confirming $f'(2)=5$.

Worked Example: A Quadratic with a Negative Leading Coefficient

Find the derivative of $g(x)=-x^2+4x-2$ at $x=-1$, directly from the definition.

Evaluate the function at $-1$:

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

Substitute the nearby input $-1+h$. Keep the entire function value in parentheses when subtracting $g(-1)$:

\begin{aligned} g'(-1) &=\lim_{h\to 0}\frac{g(-1+h)-g(-1)}{h}\\ &=\lim_{h\to 0} \frac{\bigl(-(-1+h)^2+4(-1+h)-2\bigr)-(-7)}{h}\\ &=\lim_{h\to 0} \frac{-(1-2h+h^2)-4+4h-2+7}{h}\\ &=\lim_{h\to 0}\frac{-1+2h-h^2-4+4h-2+7}{h}\\ &=\lim_{h\to 0}\frac{6h-h^2}{h}\\ &=\lim_{h\to 0}(6-h)\\ &=6. \end{aligned}

Check: the simplified quotient $6-h$ approaches $6$ as $h\to 0$. The constant terms in the numerator also sum to zero, as they must: $-1-4-2+7=0$. The result is $g'(-1)=6$.

Worked Example: A Quadratic Whose Value at the Input Is Zero

Find the derivative of $p(x)=3x^2+2x-5$ at $x=1$ using the limit definition.

At this input, the function value is zero:

$$ p(1)=3(1)^2+2(1)-5=3+2-5=0. $$

Set up and simplify the difference quotient:

\begin{aligned} p'(1) &=\lim_{h\to 0}\frac{p(1+h)-p(1)}{h}\\ &=\lim_{h\to 0} \frac{\bigl(3(1+h)^2+2(1+h)-5\bigr)-0}{h}\\ &=\lim_{h\to 0} \frac{3(1+2h+h^2)+2+2h-5}{h}\\ &=\lim_{h\to 0}\frac{3+6h+3h^2+2+2h-5}{h}\\ &=\lim_{h\to 0}\frac{8h+3h^2}{h}\\ &=\lim_{h\to 0}(8+3h)\\ &=8. \end{aligned}

Check: the quotient after cancellation is $8+3h$, and its limit as $h\to 0$ is $8$. Therefore $p'(1)=8$. Even though $p(1)=0$, the derivative is not the function value; it is the limit of the change in output divided by the change in input.

Common Pitfall: Setting the Increment to Zero Too Soon

The original difference quotient has $h$ in its denominator. If you substitute $h=0$ before simplifying, you divide by zero, so the quotient is undefined. For example, using the first worked example, direct substitution gives $0/0$, not a derivative value:

\frac{f(2+0)-f(2)}{0} = \frac{f(2)-f(2)}{0} = \frac{0}{0}.

A related mistake is to cancel terms that are added rather than factors. The correct simplification in that example is:

\frac{5h+2h^2}{h} = \frac{h(5+2h)}{h} = 5+2h \qquad (h\ne 0).

It would be incorrect to “cancel the $h$” from only one term without first recognizing that $h$ is a common factor of the entire numerator. After cancellation, take the limit. The original quotient is not evaluated at $h=0$; its simplified expression is evaluated in the limit as $h$ approaches zero.

Key takeaway: For a quadratic at a chosen input, substitute the shifted input, expand carefully, subtract the original function value, factor and cancel $h$, and only then take the limit. The constant terms must cancel before the quotient can be simplified.

Why the Calculation Matters

The method connects algebra to the geometric meaning of a derivative. Before taking the limit, the difference quotient is a secant slope. After cancellation, it is an expression whose limiting value gives the tangent slope at the chosen input. For quadratics, expansion makes that limiting value accessible using only algebra and the definition.

The same steps are useful whenever a limit-defined derivative is requested: write the quotient exactly, preserve parentheses during subtraction, expand, combine like terms, and identify a factor that can be canceled for nonzero increments. A clean calculation also makes errors easier to locate. If a constant remains in the numerator after subtracting $f(a)$, check the function evaluation and the signs in the subtraction.

Check Your Understanding

For each question, use the limit definition rather than derivative rules. Keep the increment nonzero while simplifying.

Find the derivative of $u(x)=x^2+2x+3$ at $x=1$.
For $v(x)=2x^2-x+4$, calculate $v(-1)$, then use the difference quotient to find $v'(-1)$.
Why is it invalid to substitute $h=0$ into the original difference quotient before simplifying?
In the numerator $7h-3h^2$, show the factorization that permits cancellation with a denominator of $h$.
When expanding $q(a+h)$ for a quadratic, which terms cancel after subtracting $q(a)$, and why?

Finding a Quadratic Derivative at a Point

What You'll Learn

From Secant Slopes to a Derivative at One Input

The Derivative at a Specified Input

A General Justification for Quadratics

Worked Examples

Worked Example: A Quadratic with Positive and Negative Terms

Worked Example: A Quadratic with a Negative Leading Coefficient

Worked Example: A Quadratic Whose Value at the Input Is Zero

Common Pitfall: Setting the Increment to Zero Too Soon

Why the Calculation Matters

Check Your Understanding

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Finding a Quadratic Derivative at a Point

What You'll Learn

From Secant Slopes to a Derivative at One Input

The Derivative at a Specified Input

A General Justification for Quadratics

Worked Examples

Worked Example: A Quadratic with Positive and Negative Terms

Worked Example: A Quadratic with a Negative Leading Coefficient

Worked Example: A Quadratic Whose Value at the Input Is Zero

Common Pitfall: Setting the Increment to Zero Too Soon

Why the Calculation Matters

Check Your Understanding

Test yourself: AP Calculus practice exams

Want all 1,114 practice problems?

Get the Calc Pass App

Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

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