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IBDP AA · Calculus

Four calculus mistakes—and how to catch them

A practical guide to the chain rule, stationary points, integration constants and signed area.

1. Forgetting the inside derivative

Consider f(x) = (3x² + 1)⁴. Differentiating the outside power alone gives an incomplete answer: 4(3x² + 1)³. The inside expression also changes with x.

Worked example · Chain rule
u = 3x² + 1   and   f = u⁴
df/du = 4u³   and   du/dx = 6x
f′(x) = 4(3x² + 1)³ × 6x
f′(x) = 24x(3x² + 1)³

A useful self-check is to name the inside function before differentiating. If the inside function is not simply x, ask whether its derivative has been included.

2. Treating every stationary point as a maximum or minimum

A stationary point has a zero first derivative. That condition alone does not tell you what happens on either side.

Counterexample
f(x) = x³   ⇒   f′(x) = 3x²
f′(0) = 0

For x < 0 and x > 0, the derivative is positive. The function is increasing on both sides of zero, so (0, 0) is a stationary point of inflection, not a local maximum or minimum.

Classify a stationary point by checking how the first derivative changes sign. A second derivative test can also help, but f″(a) = 0 makes that test inconclusive.

3. Losing the constant of integration

An indefinite integral describes a family of functions with the same derivative. That is why the constant matters.

∫ (6x² − 4) dx = 2x³ − 4x + C

If F(1) = 5, substitute the condition:
2 − 4 + C = 5, so C = 7.
The particular function is F(x) = 2x³ − 4x + 7.

Differentiate the result to check the integrand. Then use any stated condition to find C. A definite integral, evaluated between limits, gives a number and does not need a final “+ C”.

4. Confusing signed integral with total area

A definite integral counts regions below the x-axis negatively. Total geometric area counts every region positively.

Worked example · y = x from −1 to 1
∫−11 x dx = [x²/2]−11 = 0

The two triangular regions each have area ½. Total area is therefore ½ + ½ = 1 square unit. The signed contributions cancel in the integral; the areas do not.

Before integrating to find total area, locate where the curve crosses the x-axis and split the interval there. Reverse the sign of any negative contribution.

Your turn: apply the chain rule

Differentiate y = (2x³ − 5)⁴, then find the gradient when x = 1.

Reveal the worked solution
dy/dx = 4(2x³ − 5)³ × 6x²
= 24x²(2x³ − 5)³

At x = 1, the gradient is 24(−3)³ = −648. The factor 6x² comes from differentiating the inside expression.

For revision, keep an error log with three columns: the mistake, why it happened, and the check you will use next time. Practise the check, not only the calculation.

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