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Difference Quotient Quiz · Q1

From difference quotient to derivative — building the limit, step by step

Question 1 (1 pt) Find the general slope of the tangent line to the graph of $f(x) = 2x^2 + 1$ at any point $x$.

Choices: A. $4x$ B. $2x$ C. none of these D. $4x + 2h$

1 · The intuition

Intuition

A tangent line is a secant line whose two points have collapsed to one. So the recipe is always the same: build a secant slope between $x$ and $x + h$, then take the limit as $h \to 0$.

This problem forces you to do the whole dance — expand, cancel, factor, limit — instead of reaching for the power rule shortcut. Do it enough times and the shortcut will be the dance; the algebra just becomes a memory.

2 · The formula

Formula

For any function $f$, the derivative (general slope of the tangent line) is the limit of the difference quotient:

$$f'(x) \;=\; \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

The numerator $f(x+h) - f(x)$ is the vertical rise between the two points; the denominator $h$ is the horizontal run. Dividing gives an average slope; taking $h \to 0$ gives the instantaneous slope.

3 · Worked solution

Worked solution

Given $f(x) = 2x^2 + 1$. We compute $\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ in five steps.

Step 1 — Write the difference quotient.

$$f'(x) \;=\; \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

Step 2 — Compute $f(x+h)$ by substitution and expansion.

$$f(x+h) \;=\; 2(x+h)^2 + 1 \;=\; 2(x^2 + 2xh + h^2) + 1 \;=\; 2x^2 + 4xh + 2h^2 + 1$$

Step 3 — Subtract $f(x) = 2x^2 + 1$.

$$f(x+h) - f(x) \;=\; (2x^2 + 4xh + 2h^2 + 1) - (2x^2 + 1) \;=\; 4xh + 2h^2$$

The $2x^2$ and the $+1$ cancel — that cancellation is the whole reason this trick works.

Step 4 — Divide by $h$ and factor.

$$\frac{4xh + 2h^2}{h} \;=\; \frac{h(4x + 2h)}{h} \;=\; 4x + 2h$$

Factoring out the $h$ is what lets the $h$ cancel cleanly. If you forget to factor, the division looks stuck — and you can't take the limit yet.

Step 5 — Take the limit as $h \to 0$.

$$f'(x) \;=\; \lim_{h \to 0} (4x + 2h) \;=\; 4x + 0 \;=\; 4x$$

Final answer:

$f'(x) = 4x$

4 · Why the other choices fail

Reading the distractors

Each wrong answer maps to a specific step where someone stopped too early (or did the algebra wrong). Learning to read distractors is half of learning to do the problem.

B. $2x$ — Missed the coefficient. You treated the 2 in front of $2x^2$ as if it disappeared after one step (giving $2x^2 \to 2x$) instead of staying as a constant multiplier. If you redo the 5-step dance on $2x^2$ carefully, the leading 2 stays all the way through and ends up in the answer — that's where the 4 comes from.
C. none of these — Don't pick this without checking every option. A is real, and it's the right one.
D. $4x + 2h$ — This is the secant slope (step 4), not the tangent slope. You stopped one step too early — you never took $\lim_{h \to 0}$. The $2h$ term is what becomes the limit; the answer is what survives the limit, not what came before it.

5 · The mental model (5 steps, every time)

Pattern

Every derivative-via-the-definition problem in this course is these five steps, in this order:

Write the difference quotient with $f(x+h) - f(x)$ over $h$.
Compute $f(x+h)$ by substituting $x+h$ for $x$ and expanding.
Subtract $f(x)$ — most of the function cancels, and what remains has an $h$ in it.
Factor $h$ out of the numerator, then cancel the $h$ in the denominator.
Take the limit as $h \to 0$ — plug in $h = 0$. Whatever survives is the derivative.

If step 4 doesn't go cleanly (you can't factor $h$ out), the most common reasons are: you didn't expand $f(x+h)$ fully in step 2, or you forgot to distribute a negative in step 3.

6 · Check your understanding

Practice

Use the definition (the 5 steps above) to find $f'(x)$ for $f(x) = 3x^2 - 5x + 2$.

Hint: the constant $+2$ and the $-5x$ term will both contribute. Expand $(x+h)^2 = x^2 + 2xh + h^2$ carefully and don't lose signs when you subtract.

Show answer

Step 1: $\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Step 2: $f(x+h) = 3(x+h)^2 - 5(x+h) + 2 = 3x^2 + 6xh + 3h^2 - 5x - 5h + 2$

Step 3: $f(x+h) - f(x) = (3x^2 + 6xh + 3h^2 - 5x - 5h + 2) - (3x^2 - 5x + 2) = 6xh + 3h^2 - 5h$

Step 4: $\dfrac{h(6x + 3h - 5)}{h} = 6x + 3h - 5$

Step 5: $f'(x) = \lim_{h \to 0}(6x + 3h - 5) = 6x - 5$

$f'(x) = 6x - 5$

Sanity check via the 5-step recipe: $3x^2 \to 6x$, $-5x \to -5$, $+2 \to 0$ (constants vanish in the limit). Adds up to $6x - 5$. ✓

7 · What's next

Three directions once you send Q2:

More difference-quotient problems — keep going through the quiz, building the 5-step muscle.
Sticky questions — if any step of the dance still feels mechanical, redo Q1 with a different function like $f(x) = x^3$ or $f(x) = (x+1)^2$ to make sure the pattern generalizes.
The shortcut, later — once the 5-step dance feels automatic, the shortcut version (the power rule) will land as a "remember the dance, skip to the answer" trick. But not yet — keep doing the limit until it's muscle memory.

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