Pure mathematics

Differentiation: Advanced Techniques and Applications

Year 12 · Year 13

✓By the end of this lesson students will be able to differentiate functions from first principles.
✓By the end of this lesson students will be able to apply the product, quotient, and chain rules to differentiate complex functions.
✓By the end of this lesson students will be able to locate and classify stationary points of functions.
✓By the end of this lesson students will be able to differentiate implicitly defined functions.
✓By the end of this lesson students will be able to differentiate functions defined parametrically.

Key concepts

Differentiation from First Principles

Differentiation from first principles involves finding the derivative of a function by evaluating the limit of the gradient of a chord as the change in x (h) approaches zero. This method directly applies the definition of the derivative.

f'(x) = lim (h→0) [f(x+h) - f(x)] / h

The Product Rule

The product rule is used to differentiate a function that is the product of two other differentiable functions. If y = uv, where u and v are functions of x, then the derivative dy/dx is found by differentiating each part in turn and adding the results.

d/dx (uv) = u dv/dx + v du/dx

The Quotient Rule

The quotient rule is used to differentiate a function that is the quotient of two other differentiable functions. If y = u/v, where u and v are functions of x, then the derivative dy/dx is found using a specific formula involving the derivatives of u and v.

d/dx (u/v) = (v du/dx - u dv/dx) / v^2

The Chain Rule

The chain rule is used to differentiate composite functions, i.e., a function within a function. If y is a function of u, and u is a function of x, then the derivative dy/dx is the product of the derivative of y with respect to u and the derivative of u with respect to x.

dy/dx = dy/du × du/dx OR d/dx f(g(x)) = f'(g(x)) × g'(x)

Stationary Points

Stationary points (also known as turning points or critical points) are points on a curve where the gradient is zero, i.e., dy/dx = 0. These points can be local maxima, local minima, or points of inflection. Their nature is determined by examining the sign of the second derivative, d²y/dx².

For stationary points, dy/dx = 0. If d²y/dx² > 0, it's a local minimum. If d²y/dx² < 0, it's a local maximum. If d²y/dx² = 0, further investigation is required.

Implicit Differentiation

Implicit differentiation is a technique used to differentiate equations where y is not explicitly expressed as a function of x (e.g., x² + y² = 25). It involves differentiating both sides of the equation with respect to x, treating y as a function of x and applying the chain rule to terms involving y.

Parametric Differentiation

Parametric differentiation is used when x and y are both expressed in terms of a third variable, often 't' (the parameter). To find dy/dx, we differentiate y with respect to t (dy/dt) and x with respect to t (dx/dt), then divide dy/dt by dx/dt.

dy/dx = (dy/dt) / (dx/dt)

Key facts to remember

1The derivative f'(x) represents the gradient of the tangent to the curve y = f(x) at any point x.
2The product rule: d/dx (uv) = u dv/dx + v du/dx.
3The quotient rule: d/dx (u/v) = (v du/dx - u dv/dx) / v².
4The chain rule: dy/dx = dy/du × du/dx.
5For implicit differentiation, treat y as a function of x and apply the chain rule when differentiating terms involving y (e.g., d/dx (y²) = 2y dy/dx).
6For parametric differentiation, if x = f(t) and y = g(t), then dy/dx = (dy/dt) / (dx/dt).
7Stationary points occur where dy/dx = 0. Their nature (max/min/inflection) is determined by the sign of d²y/dx².
8Standard derivatives (e.g., d/dx (xⁿ) = nxⁿ⁻¹, d/dx (sin x) = cos x, d/dx (eˣ) = eˣ, d/dx (ln x) = 1/x) are fundamental and should be memorised.

Worked examples

Example 1

Differentiate f(x) = 2x² - 5x from first principles.

IRecall the formula for differentiation from first principles: f'(x) = lim (h→0) [f(x+h) - f(x)] / h

IISubstitute f(x+h) into the formula: f(x+h) = 2(x+h)² - 5(x+h) = 2(x² + 2xh + h²) - 5x - 5h = 2x² + 4xh + 2h² - 5x - 5h

IIISubstitute f(x) and f(x+h) into the limit expression:

IVf'(x) = lim (h→0) [(2x² + 4xh + 2h² - 5x - 5h) - (2x² - 5x)] / h

VSimplify the numerator:

VIf'(x) = lim (h→0) [2x² + 4xh + 2h² - 5x - 5h - 2x² + 5x] / h

VIIf'(x) = lim (h→0) [4xh + 2h² - 5h] / h

VIIIFactor out h from the numerator:

9f'(x) = lim (h→0) [h(4x + 2h - 5)] / h

10Cancel h (since h ≠ 0 as h approaches 0):

11f'(x) = lim (h→0) (4x + 2h - 5)

12Apply the limit as h approaches 0:

13f'(x) = 4x + 2(0) - 5

Answer

f'(x) = 4x - 5

This method demonstrates the fundamental definition of the derivative.

Example 2

Differentiate y = x² cos(3x) with respect to x.

IIdentify the two functions being multiplied: Let u = x² and v = cos(3x).

IIFind du/dx: du/dx = d/dx (x²) = 2x.

IIIFind dv/dx: This requires the chain rule. Let w = 3x, so v = cos(w). Then dv/dw = -sin(w) and dw/dx = 3. So, dv/dx = dv/dw × dw/dx = -sin(3x) × 3 = -3sin(3x).

IVApply the product rule: dy/dx = u dv/dx + v du/dx.

VSubstitute the expressions for u, v, du/dx, and dv/dx:

VIdy/dx = (x²)(-3sin(3x)) + (cos(3x))(2x)

VIIRearrange and simplify:

Answer

dy/dx = 2x cos(3x) - 3x² sin(3x)

This example combines the product rule and the chain rule, a common scenario in A-Level maths.

Example 3

Find dy/dx for the curve defined by x³ + y³ = 3xy.

IDifferentiate both sides of the equation with respect to x. Remember to apply the chain rule for terms involving y.

IId/dx (x³) + d/dx (y³) = d/dx (3xy)

IIIDifferentiate x³: d/dx (x³) = 3x².

IVDifferentiate y³: Using the chain rule, d/dx (y³) = 3y² dy/dx.

VDifferentiate 3xy: This requires the product rule. Let u = 3x and v = y. Then du/dx = 3 and dv/dx = dy/dx. So, d/dx (3xy) = (3x)(dy/dx) + (y)(3) = 3x dy/dx + 3y.

VISubstitute these derivatives back into the equation:

VII3x² + 3y² dy/dx = 3x dy/dx + 3y

VIIIRearrange the equation to collect all terms involving dy/dx on one side:

93y² dy/dx - 3x dy/dx = 3y - 3x²

10Factor out dy/dx:

11dy/dx (3y² - 3x) = 3y - 3x²

12Isolate dy/dx:

Answer

dy/dx = (3y - 3x²) / (3y² - 3x) = (y - x²) / (y² - x)

Implicit differentiation is crucial for curves where y cannot be easily expressed as a function of x.

Common mistakes

✗Forgetting to apply the chain rule when differentiating composite functions or terms involving y in implicit differentiation.
✗Making sign errors or incorrect order of terms in the quotient rule formula.
✗Algebraic errors when simplifying expressions after applying differentiation rules.
✗Incorrectly differentiating constants or terms that do not involve the variable of differentiation.
✗Failing to factorise dy/dx correctly when solving for it in implicit differentiation problems.

Exam tips

★Always show full working, especially for differentiation from first principles, as method marks are crucial.
★Clearly identify which differentiation rule(s) you are using before you start, particularly for complex functions.
★Pay close attention to notation (e.g., dy/dx, f'(x)) and ensure it is used consistently and correctly.
★Simplify your final answer where possible, but be careful not to introduce errors during simplification.

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