Common Mistakes

Traps in Application of Derivatives

6 mistake patterns students fall for. 2 high-frequency traps appear in almost every exam.

Confusing local extrema with global extrema

Very CommonCASE MISS

A local maximum is the largest value in a neighbourhood, not necessarily on the entire domain. The global max on [a, b] requires checking endpoints too.

Why: Students find f'(x) = 0 and assume the resulting value is the absolute max or min without comparing with endpoint values.

WRONG: Finding f'(x) = 0 at x = c and declaring f(c) as the maximum on [a, b] without evaluating f(a) and f(b)
RIGHT: Evaluate f at all critical points AND at both endpoints. The global max = max{f(a), f(c1), f(c2), ..., f(b)}.
See pattern: Absolute Max/Min on Closed Interval

Missing critical points where f'(x) does not exist

Very CommonDOMAIN

Critical points include both where f'(x) = 0 and where f'(x) is undefined (cusps, corners, vertical tangents). Missing the latter leads to incomplete analysis.

Why: Students only solve f'(x) = 0 and forget to check points where the derivative fails to exist, such as |x| at x = 0.

WRONG: For f(x) = |x - 2|, only solving f'(x) = 0 and finding no solution, concluding no critical points
RIGHT: f'(x) does not exist at x = 2 (corner point). So x = 2 is a critical point. Always check differentiability of the function.
See pattern: Local Maxima and Minima

Wrong sign in normal equation slope

CommonSIGN ERROR

The slope of the normal is -1/f'(x1), not 1/f'(x1). Forgetting the negative sign gives a line that is not perpendicular to the tangent.

Why: Students remember that the normal slope involves the reciprocal but forget the negative sign required for perpendicularity.

WRONG: Tangent slope = 3, so normal slope = 1/3
RIGHT: Tangent slope = 3, so normal slope = -1/3. Perpendicular lines have slopes whose product is -1.
See pattern: Tangent and Normal Equations

Not checking endpoints for absolute max/min

CommonCASE MISS

On a closed interval [a, b], the absolute extremum can occur at an endpoint, not just at interior critical points.

Why: Students focus entirely on solving f'(x) = 0 and forget that the closed interval theorem requires evaluating f at the boundary.

WRONG: Finding critical point at x = 2 in [0, 5] and reporting f(2) as the answer without checking f(0) and f(5)
RIGHT: Compute f(0), f(2), and f(5). Compare all three values. The largest is the absolute max and the smallest is the absolute min.
See pattern: Absolute Max/Min on Closed Interval

Applying second derivative test when f''(c) = 0

OccasionalFORMULA

The second derivative test is inconclusive when f''(c) = 0. Students incorrectly conclude it is a point of inflection or apply the test anyway.

Why: Students memorize f''(c) > 0 means min and f''(c) < 0 means max, but do not learn the inconclusive case f''(c) = 0.

WRONG: f'(0) = 0, f''(0) = 0, so x = 0 is a point of inflection (wrong for f(x) = x^4, which has a minimum at 0)
RIGHT: When f''(c) = 0, use the first derivative test. Check sign change of f'(x) around x = c to determine the nature of the critical point.
See pattern: Local Maxima and Minima

Forgetting to verify Rolle's/LMVT conditions

OccasionalCASE MISS

Rolle's theorem requires continuity on [a, b], differentiability on (a, b), and f(a) = f(b). Applying it without checking all three conditions leads to invalid conclusions.

Why: Students jump directly to finding c where f'(c) = 0 without verifying that the hypothesis of the theorem is satisfied.

WRONG: Applying Rolle's theorem to f(x) = |x| on [-1, 1] without noting that f'(0) does not exist (not differentiable on (-1, 1))
RIGHT: First verify: (1) f is continuous on [a, b], (2) f is differentiable on (a, b), (3) f(a) = f(b). Only then conclude that there exists c in (a, b) with f'(c) = 0.
See pattern: Rolle's Theorem and LMVT Applications
Test yourself

Can you spot these traps under time pressure?

Take a timed quiz on Application of Derivatives and see if you avoid the mistakes above.