Limits
Always substitute the value first. If you get a finite number, that is the limit. Only an indeterminate form - most often 0/0 - requires any technique.
- ✓Standard limits worth memorising: lim(x→0) sinx/x = 1; lim(x→0) tanx/x = 1; lim(x→0) (1 − cosx)/x² = ½; lim(x→0) (eˣ − 1)/x = 1; lim(x→0) ln(1+x)/x = 1; lim(x→a) (xⁿ − aⁿ)/(x − a) = naⁿ⁻¹.
Evaluate lim(x→2) (x² − 4)/(x − 2).
4. Direct substitution gives 0/0, so factorise: (x−2)(x+2)/(x−2) = x + 2, which tends to 4. Cancelling is legitimate because x approaches 2 without ever equalling it.
Evaluate lim(x→0) sin3x/x.
3. Rewrite as 3 × (sin3x/3x); as x → 0, the bracket tends to 1 by the standard limit, leaving 3. In general lim(x→0) sin(kx)/x = k.
Evaluate lim(x→0) (√(1+x) − 1)/x.
½. Rationalise by multiplying by the conjugate: the numerator becomes (1 + x) − 1 = x, giving x/[x(√(1+x) + 1)] = 1/(√(1+x) + 1), which tends to 1/2.
Is f(x) = |x| differentiable at x = 0?
It is continuous there but not differentiable. The left-hand derivative is −1 and the right-hand derivative is +1, so no single tangent exists at the corner. Continuity does not imply differentiability, though differentiability does imply continuity.
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Differentiation
- ✓The table: d/dx(xⁿ) = nxⁿ⁻¹; sinx → cosx; cosx → −sinx; tanx → sec²x; eˣ → eˣ; ln x → 1/x; aˣ → aˣ ln a.
- ✓Product rule: (uv)′ = u′v + uv′. Quotient rule: (u/v)′ = (u′v − uv′)/v². Chain rule: dy/dx = (dy/du)(du/dx).
Differentiate y = x² sin x.
dy/dx = 2x sin x + x² cos x, by the product rule with u = x² and v = sin x. Neither term may be dropped - forgetting one is the standard error here.
Differentiate y = (3x + 1)⁵.
dy/dx = 15(3x + 1)⁴. By the chain rule: differentiate the outer power to get 5(3x+1)⁴, then multiply by the derivative of the inside, 3. Omitting that inner derivative is the most common chain-rule mistake.
Differentiate y = ln(x² + 1).
dy/dx = 2x/(x² + 1). The derivative of ln u is u′/u, which is the chain rule applied to the logarithm.
Differentiate y = x/(x + 1).
dy/dx = 1/(x + 1)². By the quotient rule: [(1)(x+1) − (x)(1)]/(x+1)² = 1/(x+1)². Remember the order in the numerator - u′v minus uv′, not the reverse.
Find the slope of the tangent to y = x² at x = 3, and the equation of that tangent.
dy/dx = 2x = 6 at x = 3, where y = 9. The tangent is y − 9 = 6(x − 3), that is y = 6x − 9. The normal would have slope −1/6.
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Maxima, minima and rates of change
Find the maximum value of f(x) = −x² + 4x + 1.
5, at x = 2. Set f′(x) = −2x + 4 = 0 to get x = 2; since f″(x) = −2 < 0 this is a maximum, and f(2) = −4 + 8 + 1 = 5. Always confirm the nature with the second derivative rather than assuming.
A balloon's radius increases at 2 cm/s. How fast is its volume increasing when r = 5 cm?
200π cm³/s. Differentiate V = (4/3)πr³ with respect to time: dV/dt = 4πr² (dr/dt) = 4π × 25 × 2. Related-rates problems are just the chain rule applied through time.
Divide 20 into two parts whose product is maximum.
10 and 10, giving 100. Let the parts be x and 20 − x, so P = 20x − x²; P′ = 20 − 2x = 0 gives x = 10, and P″ = −2 < 0 confirms a maximum. The two-equal-parts result generalises to any fixed sum.
For what values of x is f(x) = x³ − 3x increasing?
For x < −1 and x > 1. f′(x) = 3x² − 3 = 3(x−1)(x+1), which is positive outside [−1, 1]. The function decreases on (−1, 1), with a local maximum at x = −1 and a local minimum at x = 1.
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Integration
- ✓Standard forms: ∫xⁿ dx = xⁿ⁺¹/(n+1) + C for n ≠ −1; ∫dx/x = ln|x| + C; ∫eˣ dx = eˣ + C; ∫sinx dx = −cosx + C; ∫cosx dx = sinx + C; ∫sec²x dx = tanx + C.
- ✓By parts: ∫u dv = uv − ∫v du, choosing u by the ILATE order (inverse, log, algebraic, trigonometric, exponential).
Evaluate ∫(2x + 3) dx.
x² + 3x + C. Integrate term by term. The constant of integration is not optional in an indefinite integral - omitting it loses a mark.
Evaluate ∫₀¹ x² dx.
1/3. The antiderivative is x³/3, and evaluating from 0 to 1 gives 1/3 − 0. Definite integrals need no constant, since it cancels between the limits.
Evaluate ∫ sin²x dx.
x/2 − (sin 2x)/4 + C. Use the identity sin²x = (1 − cos 2x)/2 first; integrating gives x/2 − (sin 2x)/4. Squared trigonometric functions almost always need a double-angle identity before integration.
Evaluate ∫ x eˣ dx.
x eˣ − eˣ + C = eˣ(x − 1) + C. Integrate by parts with u = x (algebraic, so it comes before exponential in ILATE) and dv = eˣ dx, giving x eˣ − ∫eˣ dx.
Evaluate ∫₀^(π/2) sin x dx.
1. The antiderivative is −cos x, and (−cos(π/2)) − (−cos 0) = 0 + 1 = 1. Watch the sign - the derivative of cos is negative sin, so the integral of sin carries a minus.
Find the area bounded by y = x, the x-axis, and the lines x = 0 and x = 2.
2 square units. Area = ∫₀² x dx = [x²/2]₀² = 2. Here the geometry confirms it: the region is a triangle of base 2 and height 2.
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Where calculus marks are lost
- ✓Skipping the inner derivative in the chain rule - the single most common differentiation error.
- ✓Omitting the constant of integration in an indefinite integral.
- ✓Applying L'Hôpital's rule to a form that is not 0/0 or ∞/∞. Check the form before differentiating.
- ✓Assuming a stationary point is a maximum without testing the second derivative.
- ✓Forgetting that continuity does not imply differentiability - corners and cusps are continuous but have no tangent.
- ✓Integrating squared trigonometric functions directly instead of converting with a double-angle identity first.
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