Differentiation and Its Applications

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1Summary 重點摘要

The derivative is the instantaneous rate of change, numerically the tangent slope; master the power, trigonometric, exponential and logarithmic formulas plus the sum, product, quotient and chain rules, then use the second derivative to classify extrema and concavity — ready for tangents, curve sketching and rate problems.

2Vocabulary cards Vocabulary

3Step by step Steps

1. Step 1: The definition and first principlesLet the h in the secant slope [f(x+h)−f(x)]/h tend to zero; the resulting limit is the derivative f'(x) = lim h→0 [f(x+h)−f(x)]/h — differentiation from first principles. For example, for f(x) = x², [f(x+h)−f(x)]/h = (2xh+h²)/h = 2x+h, and letting h → 0 gives f'(x) = 2x. The curriculum requires finding, from first principles, the derivatives of C, x to the n, √x, sin x, cos x, e to the x and ln x. The notations y', f'(x) and dy/dx run in parallel; at a point we write f'(x₀) or dy/dx evaluated at x = x₀.
2. Step 2: The basic differentiation formulasThe formula table must be second nature: (C)' = 0; (xⁿ)' = nxⁿ⁻¹; (sin x)' = cos x; (cos x)' = −sin x; (tan x)' = sec²x; (eˣ)' = eˣ; (ln x)' = 1/x. For example, the derivative of y = x⁵ is 5x⁴; for y = √x = x^½ the derivative is (12)x^(−12), i.e. 1/(2√x). Whenever a surd or a fraction appears, rewrite it as a power before applying the power rule — this single habit prevents most computational slips. And remember: the derivative of a constant is zero.
3. Step 3: The sum ruleThe derivative of a sum or difference is the sum or difference of the derivatives, and constant factors factor out: d(u±v)/dx = du/dx ± dv/dx. Handle polynomials term by term. For example, y = 4x³ − 3x² + 5x − 7 differentiates term by term to y' = 12x² − 6x + 5, the constant term 7 giving zero. This rule breaks long expressions into manageable pieces and is the first step of every computation; the real challenges are the product, quotient and chain rules, but term-by-term thinking remains the basis for organising any expression.
4. Step 4: The product ruleThe derivative of a product is not the product of derivatives: d(uv)/dx = v(du/dx) + u(dv/dx) — the derivative of the first times the second, plus the first times the derivative of the second. For example, y = x² eˣ: with u = x², v = eˣ, we get y' = eˣ(2x) + x²(eˣ) = eˣ(x² + 2x). Writing u and v down explicitly before substituting into the formula greatly reduces missed terms; whenever two different types of function are multiplied — polynomial times exponential, x times logarithm — the product rule should come to mind immediately.
5. Step 5: The quotient ruled(u/v)/dx = [v(du/dx) − u(dv/dx)]/v²: the numerator is the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, all over the square of the denominator; the order must not be swapped. For example, y = x/(2x+1): y' = [(2x+1)(1) − x(2)]/(2x+1)² = 1/(2x+1)². Swapping the two numerator terms flips the sign and ruins the whole question; the habit of writing 'denominator times top derivative' first, then 'minus numerator times bottom derivative', plus a quick check that the first numerator term matches the denominator's type, catches order errors early.
6. Step 6: The chain ruleFor the composite function y = f(g(x)), the derivative is dy/dx = (dy/du)(du/dx): differentiate the outer function with respect to the inner, then multiply by the derivative of the inner. For example, y = (3x+1)⁵: with u = 3x+1, y = u⁵, dy/du = 5u⁴ and du/dx = 3, so y' = 15(3x+1)⁴. Whenever a function hides inside a root, a logarithm or a trigonometric sign, remember to multiply by the inner derivative; omitting it is the most common error. Another example: y = ln(x²+1) differentiates to 2x/(x²+1) — the outer layer gives 1/(x²+1), then multiply by the inner 2x.
7. Step 7: Implicit differentiation and the second derivativeFor equations that cannot be written as y = f(x), differentiate both sides with respect to x, attaching dy/dx to every y-term, then solve for dy/dx — implicit differentiation, which includes logarithmic differentiation. Differentiating again gives the second derivative f''(x), also written d²y/dx². For example, y = x³ gives y' = 3x² and y'' = 6x. The second derivative test: at a stationary point, f'' > 0 means a local minimum and f'' < 0 a local maximum; the sign of f'' also gives concavity, upward or downward. The curriculum explicitly excludes third and higher derivatives.
8. Step 8: Applications: tangents, extrema and ratesThree main applications. First, tangent equations: the slope is f'(x₀), through the point (x₀, f(x₀)), using the point-slope form y − f(x₀) = f'(x₀)(x − x₀). Second, maxima and minima: solve f'(x) = 0 for stationary points, classify with the second derivative or a sign chart, and distinguish global from local extrema, checking the endpoints of the domain. Third, rate problems: translate the wording into a function and differentiate. For example, the tangent to y = x² at x = 3 is y = 6x − 9; the derivative of a balloon's radius with respect to time is its inflation rate — both are staples of DSE long questions.

5Worked example Example

6Common mistakes Mistakes

7Quick check Quick check

8Recap Recap

• The derivative is the instantaneous rate of change: f'(x) = lim h→0 [f(x+h)−f(x)]/h, geometrically the tangent slope.
• The formula table plus the four rules (sum, product, quotient, chain) are all the tools of differentiation; convert surds and fractions to powers first.
• Product and quotient rules have fixed structures — write u and v before substituting; never forget the inner derivative in the chain rule.
• Extrema in three moves: solve f'(x) = 0 for stationary points, classify with f'', and check the domain for global extrema.
• Tangents use the point-slope form y − f(a) = f'(a)(x − a); for rate problems, define the variables, write the function, then differentiate.