Math · Calculus I · Worked example
Find a derivative from the limit definition
Use the limit definition to find the slope of f(x) = x² at x = 2, then write the tangent line there.
Write the difference quotient at a = 2
Replace f(2 + h) by (2 + h)² and f(2) by 2² = 4. This quotient is the slope of the secant between x = 2 and x = 2 + h.
Expand the numerator
(2 + h)² = 4 + 4h + h². Subtracting 4 cancels the constant term, which is what always happens when f is continuous at the point.
Cancel h, keeping h ≠ 0
Factor h out of the numerator and cancel it. The cancellation is valid because the quotient is only ever used at nonzero h; the simplified form 4 + h is what the secant slopes look like near h = 0.
Take the limit
As h approaches 0 from either side, 4 + h approaches 4. So the tangent slope at x = 2 is f′(2) = 4.
Compare with nearby secants
The secant slopes 4 + h sit just above 4 for small positive h and just below it for small negative h. The table supports the limit; the algebra is what establishes it.
| h | Secant slope 4 + h |
|---|---|
| 0.1 | 4.1 |
| 0.01 | 4.01 |
| −0.01 | 3.99 |
| −0.1 | 3.9 |
Write the tangent line
The tangent passes through (2, f(2)) = (2, 4) with slope 4. Point-slope form, y = f(a) + f′(a)(x − a), gives the line; expanding writes it as y = 4x − 4.
Result
f′(2) = 4, and the tangent line at x = 2 is y = 4x − 4.
Your turn
Use the limit definition to find the tangent line to f(x) = x² at x = −1.
Show the answer and explanation
y = −2x − 1.
The quotient ((−1 + h)² − 1)/h = (−2h + h²)/h = −2 + h for h ≠ 0, which approaches −2, so f′(−1) = −2. The point is (−1, 1), so the tangent is y = 1 − 2(x + 1) = −2x − 1.
Keep exploring
Open Secant → Tangent at x = 2 and drag h toward zero: the secant slope 4 + h settles at 4.
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