5.6

The Chain Rule (Composite Functions)

  • Chain Rule Formula: For composite function $y = f(g(x))$, let $u = g(x)$: $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \quad \text{or} \quad \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$$
  • Generalized Power Rule: $$\frac{d}{dx}\left([u(x)]^n\right) = n [u(x)]^{n-1} \cdot u'(x)$$
  • Transcendental Chain Rule Derivatives: $$\frac{d}{dx}\left(e^{u(x)}\right) = e^{u(x)} \cdot u'(x), \quad \frac{d}{dx}(\ln(u(x))) = \frac{u'(x)}{u(x)}$$ $$\frac{d}{dx}(\sin(u(x))) = \cos(u(x)) \cdot u'(x), \quad \frac{d}{dx}(\cos(u(x))) = -\sin(u(x)) \cdot u'(x)$$

Example 1

Differentiating Composite Functions

Differentiate each function with respect to $x$:
(a) $y = (3x^2 - 5x + 2)^4$
(b) $f(x) = \sqrt{9 - x^2}$
(c) $g(x) = e^{\sin(2x)}$

  1. (a) Outer function: $(\dots)^4$; Inner function: $u = 3x^2 - 5x + 2 \implies u' = 6x - 5$: $$\frac{dy}{dx} = 4(3x^2 - 5x + 2)^3 \cdot (6x - 5) = \mathbf{4(6x - 5)(3x^2 - 5x + 2)^3}$$
  2. (b) Rewrite as power: $f(x) = (9 - x^2)^{1/2}$: $$f'(x) = \frac{1}{2}(9 - x^2)^{-1/2} \cdot (-2x) = \mathbf{-\frac{x}{\sqrt{9 - x^2}}}$$
  3. (c) Inner function $u = \sin(2x) \implies u' = 2\cos(2x)$: $$g'(x) = e^{\sin(2x)} \cdot (2\cos(2x)) = \mathbf{2\cos(2x)e^{\sin(2x)}}$$

Answers: (a) $4(6x - 5)(3x^2 - 5x + 2)^3$; (b) $-\frac{x}{\sqrt{9 - x^2}}$; (c) $2\cos(2x)e^{\sin(2x)}$.

5.6

The Product Rule (Multiplying Functions)

  • Product Rule Formula: When differentiating the product of two differentiable functions $y = u \cdot v$: $$\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \quad \text{or} \quad (uv)' = u'v + uv'$$
  • When to Use: When two separate expressions containing $x$ are multiplied together and cannot be easily expanded into single polynomial terms.
  • Exam habit: Explicitly declare $u = \dots$ and $v = \dots$ on your exam paper, find $u'$ and $v'$ separately, then cross-multiply and factorize the common terms.

Example 2

Product Rule with Factorization

Differentiate $y = x^3 e^{2x}$ and express your answer in fully factorized form.

  1. Identify components: $$u = x^3 \implies u' = 3x^2$$ $$v = e^{2x} \implies v' = 2e^{2x} \quad (\text{by Chain Rule})$$
  2. Apply Product Rule: $$\frac{dy}{dx} = u'v + uv' = (3x^2)(e^{2x}) + (x^3)(2e^{2x})$$
  3. Factorize common factors ($x^2 e^{2x}$): $$\frac{dy}{dx} = x^2 e^{2x}(3 + 2x) = \mathbf{x^2(2x + 3)e^{2x}}$$

Answer: $\frac{dy}{dx} = x^2(2x + 3)e^{2x}$.

5.6

The Quotient Rule (Fractions of Functions)

  • Quotient Rule Formula: For function $y = \frac{u}{v}$ where $v(x) \ne 0$: $$\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \quad \text{or} \quad \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$$
  • Critical Sign Alert: Unlike the product rule, the numerator has a subtraction sign: $u'v - uv'$. Starting with the wrong term ($uv' - u'v$) causes an immediate sign error across your entire solution.
  • Denominator Tip: Leave the denominator $(v)^2$ in factored form—never expand it unless explicitly required to simplify the whole fraction.

Example 3

Differentiating with the Quotient Rule

Find the derivative of $f(x) = \frac{\ln(x)}{x^2}$ for $x > 0$.

  1. Identify numerator and denominator: $$u = \ln(x) \implies u' = \frac{1}{x}$$ $$v = x^2 \implies v' = 2x$$
  2. Apply Quotient Rule: $$f'(x) = \frac{u'v - uv'}{v^2} = \frac{\left(\frac{1}{x}\right)(x^2) - (\ln(x))(2x)}{(x^2)^2}$$
  3. Simplify numerator: $$f'(x) = \frac{x - 2x\ln(x)}{x^4} = \frac{x(1 - 2\ln(x))}{x^4} = \mathbf{\frac{1 - 2\ln(x)}{x^3}}$$

Answer: $f'(x) = \frac{1 - 2\ln(x)}{x^3}$.

5.4

Tangent Lines to Advanced Curves

  • Tangent Line Equation: The tangent line touches the curve $y = f(x)$ at point $P(x_1, y_1)$ with matching slope: $$m_T = f'(x_1)$$ $$y - y_1 = m_T(x - x_1)$$
  • Systematic Steps:
    1. If $y_1$ is not given, find it by evaluating $y_1 = f(x_1)$.
    2. Differentiate $f(x)$ using appropriate rule (chain, product, quotient).
    3. Evaluate gradient $m_T = f'(x_1)$.
    4. Form point-slope equation and rearrange into requested form ($y = mx + c$ or $ax + by + d = 0$).

Example 4

Finding Tangent Equation Using Product Rule

Find the equation of the tangent to the curve $y = (2x - 1)e^x$ at $x = 0$. Give your answer in the form $y = mx + c$.

  1. Find point on curve: $$\text{At } x = 0: \quad y_1 = (2(0) - 1)e^0 = (-1)(1) = -1 \implies (0, -1)$$
  2. Differentiate using Product Rule: $$\frac{dy}{dx} = (2)(e^x) + (2x - 1)(e^x) = e^x(2 + 2x - 1) = (2x + 1)e^x$$
  3. Find gradient at $x = 0$: $$m_T = (2(0) + 1)e^0 = 1(1) = 1$$
  4. Write line equation: $$y - (-1) = 1(x - 0) \implies y + 1 = x \implies \mathbf{y = x - 1}$$

Answer: $y = x - 1$.

5.4

Normal Lines (Perpendicular to Tangent)

  • Normal Line Definition: The line perpendicular to the tangent at the point of contact $P(x_1, y_1)$.
  • Perpendicular Gradient Rule: $$m_T \cdot m_N = -1 \implies m_N = -\frac{1}{m_T} = -\frac{1}{f'(x_1)}$$
  • Special Cases:
    • If tangent is horizontal ($m_T = 0$), the normal is vertical: $x = x_1$.
    • If tangent is vertical ($m_T$ undefined), the normal is horizontal: $y = y_1$.
Tangent and Normal Lines Geometry Illustrates a curve with a solid tangent line touching at point P and a dashed normal line perpendicular to the tangent with right-angle indicator. x y y = f(x) Tangent: slope = m Normal: slope = −1 / m P(x₁, y₁)
Figure 5.5: Tangent and normal lines to a curve at $P(x_1, y_1)$ are perpendicular: $m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}}$.

Example 5

Finding the Normal Line Equation

Find the equation of the normal to the curve $y = \frac{4}{2x - 1}$ at the point where $x = 1.5$. Give your answer in the form $ax + by + d = 0$ where $a, b, d \in \mathbb{Z}$.

  1. Find $y$-coordinate at $x = 1.5$: $$y = \frac{4}{2(1.5) - 1} = \frac{4}{3 - 1} = \frac{4}{2} = 2 \implies P(1.5, 2)$$
  2. Differentiate using Chain Rule ($y = 4(2x - 1)^{-1}$): $$\frac{dy}{dx} = 4(-1)(2x - 1)^{-2}(2) = -\frac{8}{(2x - 1)^2}$$
  3. Evaluate tangent gradient at $x = 1.5$: $$m_T = -\frac{8}{(2(1.5) - 1)^2} = -\frac{8}{2^2} = -\frac{8}{4} = -2$$
  4. Find normal gradient: $$m_N = -\frac{1}{m_T} = -\frac{1}{-2} = \frac{1}{2}$$
  5. Write normal equation: $$y - 2 = \frac{1}{2}(x - 1.5) \implies 2(y - 2) = x - 1.5 \implies 2y - 4 = x - 1.5$$ Multiply by 2 to clear fractions: $$4y - 8 = 2x - 3 \implies \mathbf{2x - 4y + 5 = 0}$$

Answer: $2x - 4y + 5 = 0$.

Tier 1: Core Differentiation Rules (Questions 1–4)

  1. 1. Chain Rule with Radicals: Differentiate $f(x) = \sqrt{2x^3 + 5}$ with respect to $x$.

    Show solution

    Rewrite as power: $f(x) = (2x^3 + 5)^{1/2}$.
    Outer derivative: $\frac{1}{2}(2x^3 + 5)^{-1/2}$.
    Inner derivative: $\frac{d}{dx}(2x^3 + 5) = 6x^2$.
    $$f'(x) = \frac{1}{2}(2x^3 + 5)^{-1/2} \cdot 6x^2 = \mathbf{\frac{3x^2}{\sqrt{2x^3 + 5}}}$$

  2. 2. Product Rule with Trigonometry: Differentiate $y = x^2 \cos(3x)$.

    Show solution

    Let $u = x^2 \implies u' = 2x$.
    Let $v = \cos(3x) \implies v' = -3\sin(3x)$ (by Chain Rule).
    $$\frac{dy}{dx} = u'v + uv' = (2x)(\cos(3x)) + (x^2)(-3\sin(3x)) = \mathbf{2x\cos(3x) - 3x^2\sin(3x)}$$

  3. 3. Quotient Rule: Differentiate $g(x) = \frac{3x - 2}{x^2 + 1}$.

    Show solution

    $u = 3x - 2 \implies u' = 3$.
    $v = x^2 + 1 \implies v' = 2x$.
    $$g'(x) = \frac{u'v - uv'}{v^2} = \frac{3(x^2 + 1) - (3x - 2)(2x)}{(x^2 + 1)^2}$$ $$g'(x) = \frac{3x^2 + 3 - (6x^2 - 4x)}{(x^2 + 1)^2} = \mathbf{\frac{-3x^2 + 4x + 3}{(x^2 + 1)^2}}$$

  4. 4. Normal Gradient: The gradient of the tangent to a curve at $x = 3$ is $-\frac{3}{5}$. State the gradient of the normal line at this point.

    Show solution

    The normal line is perpendicular to the tangent line: $$m_N = -\frac{1}{m_T} = -\frac{1}{-3/5} = \mathbf{\frac{5}{3}}$$

Tier 2: Tangent & Normal Modeling (Questions 5–7)

  1. 5. Complete Tangent Equation: Find the equation of the tangent line to the curve $y = x^2 \ln(x)$ at $x = e$. Give your answer in the form $y = mx + c$.

    Show solution

    Step 1: Find $y$-coordinate at $x = e$: $$y = (e)^2 \ln(e) = e^2(1) = e^2 \implies (e, e^2)$$ Step 2: Differentiate using Product Rule: $$\frac{dy}{dx} = (2x)(\ln(x)) + (x^2)\left(\frac{1}{x}\right) = 2x\ln(x) + x$$ Step 3: Evaluate gradient at $x = e$: $$m_T = 2e\ln(e) + e = 2e(1) + e = 3e$$ Step 4: Tangent line equation: $$y - e^2 = 3e(x - e) \implies y - e^2 = 3ex - 3e^2 \implies \mathbf{y = 3ex - 2e^2}$$

  2. 6. Normal Line to an Exponential Curve: Find the equation of the normal to the curve $y = 3e^{2x} - 1$ at the point where $x = 0$.

    Show solution

    Step 1: Point coordinates: $$\text{At } x = 0: \quad y = 3e^0 - 1 = 3(1) - 1 = 2 \implies (0, 2)$$ Step 2: Differentiate: $$\frac{dy}{dx} = 3(2e^{2x}) = 6e^{2x}$$ $$\text{At } x = 0: \quad m_T = 6e^0 = 6$$ Step 3: Normal gradient: $$m_N = -\frac{1}{6}$$ Step 4: Equation of normal: $$y - 2 = -\frac{1}{6}(x - 0) \implies \mathbf{y = -\frac{1}{6}x + 2} \quad \text{or} \quad \mathbf{x + 6y - 12 = 0}$$

  3. 7. Tangent Parallel to a Given Line: Find the coordinates of the point on the curve $y = \frac{x}{x + 2}$ where the tangent is parallel to the line $y = 2x + 1$.

    Show solution

    Step 1: Differentiate $y = \frac{x}{x+2}$ using Quotient Rule: $$\frac{dy}{dx} = \frac{(1)(x + 2) - (x)(1)}{(x + 2)^2} = \frac{2}{(x + 2)^2}$$ Step 2: Set $\frac{dy}{dx} = 2$ (since parallel to line with gradient $2$): $$\frac{2}{(x + 2)^2} = 2 \implies (x + 2)^2 = 1 \implies x + 2 = \pm 1$$ $$x = -1 \quad \text{or} \quad x = -3$$ Step 3: Find corresponding $y$-values: $$\text{At } x = -1: \quad y = \frac{-1}{-1 + 2} = -1 \implies \mathbf{(-1, -1)}$$ $$\text{At } x = -3: \quad y = \frac{-3}{-3 + 2} = \frac{-3}{-1} = 3 \implies \mathbf{(-3, 3)}$$

Tier 3: Complex Applications & Intersections (Questions 8–10)

  1. 8. Normal Intersecting the Curve Again: The normal to the curve $y = x^2 - 4x + 3$ at the point $P(1, 0)$ intersects the curve again at point $Q$. Find the coordinates of $Q$.

    Show solution

    Step 1: Tangent and normal gradient at $P(1, 0)$: $$\frac{dy}{dx} = 2x - 4 \implies m_T = 2(1) - 4 = -2$$ $$m_N = -\frac{1}{-2} = \frac{1}{2}$$ Step 2: Normal equation: $$y - 0 = \frac{1}{2}(x - 1) \implies y = \frac{1}{2}x - \frac{1}{2}$$ Step 3: Intersect with curve $y = x^2 - 4x + 3$: $$x^2 - 4x + 3 = \frac{1}{2}x - \frac{1}{2} \implies 2x^2 - 8x + 6 = x - 1 \implies 2x^2 - 9x + 7 = 0$$ $$(2x - 7)(x - 1) = 0 \implies x = 1 \text{ (point } P) \quad \text{or} \quad x = \frac{7}{2} = 3.5$$ Step 4: Find $y$-coordinate of $Q$: $$y = \frac{1}{2}(3.5) - 0.5 = 1.75 - 0.5 = 1.25 \implies \mathbf{Q(3.5, 1.25)} \quad \text{or} \quad \mathbf{Q\left(\frac{7}{2}, \frac{5}{4}\right)}$$

  2. 9. Tangent to Trigonometric Curve: Consider the curve $y = 2\sin(x) + \cos(2x)$ for $0 \le x \le \pi$.
    (a) Find $\frac{dy}{dx}$.
    (b) Find the equation of the tangent line at $x = \frac{\pi}{6}$.

    Show solution

    (a) Differentiate: $$\frac{dy}{dx} = 2\cos(x) - 2\sin(2x)$$ (b) At $x = \frac{\pi}{6}$: $$y = 2\sin\left(\frac{\pi}{6}\right) + \cos\left(\frac{\pi}{3}\right) = 2(0.5) + 0.5 = 1.5 = \frac{3}{2}$$ $$m_T = 2\cos\left(\frac{\pi}{6}\right) - 2\sin\left(\frac{\pi}{3}\right) = 2\left(\frac{\sqrt{3}}{2}\right) - 2\left(\frac{\sqrt{3}}{2}\right) = 0$$ Since $m_T = 0$, the tangent line is horizontal: $$\mathbf{y = \frac{3}{2}}$$

  3. 10. Intersection of Tangent and Normal with Coordinate Axes: The curve $y = e^{2x}$ has a tangent and a normal drawn at the point $P(0, 1)$.
    (a) Find the equations of the tangent and normal lines.
    (b) The tangent intersects the $x$-axis at $A$, and the normal intersects the $x$-axis at $B$. Calculate the area of triangle $PAB$.

    Show solution

    (a) $\frac{dy}{dx} = 2e^{2x} \implies m_T = 2e^0 = 2$.
    Tangent equation: $y - 1 = 2(x - 0) \implies \mathbf{y = 2x + 1}$.
    Normal gradient: $m_N = -\frac{1}{2}$.
    Normal equation: $y - 1 = -\frac{1}{2}(x - 0) \implies \mathbf{y = -\frac{1}{2}x + 1}$.

    (b) Find $x$-intercepts ($y = 0$):
    Tangent: $0 = 2x + 1 \implies x = -0.5 \implies A(-0.5, 0)$.
    Normal: $0 = -\frac{1}{2}x + 1 \implies x = 2 \implies B(2, 0)$.
    Base $AB = 2 - (-0.5) = 2.5$.
    Height from $P(0, 1)$ to $x$-axis is $1$.
    $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2.5 \times 1 = \mathbf{1.25\text{ units}^2}$$

Chain Rule

Formula Booklet
$$y = g(u), u = f(x) \implies \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$$

Differentiate outer function then multiply by derivative of inside function.

Product Rule

Formula Booklet
$$y = uv \implies \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}$$

Sum of cross-derivatives when two functions of $x$ are multiplied.

Quotient Rule

Formula Booklet
$$y = \frac{u}{v} \implies \frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$$

Numerator order matters: $u'v - uv'$. Leave denominator factored.

Tangent Line Equation

Memorize
$$y - y_1 = m_T(x - x_1), \quad m_T = f'(x_1)$$

Point-slope formula with derivative gradient at the contact point.

Normal Line Equation

Memorize
$$m_N = -\frac{1}{m_T} \implies y - y_1 = -\frac{1}{f'(x_1)}(x - x_1)$$

Perpendicular gradient condition $m_1 m_2 = -1$. Vertical if $m_T = 0$.

Exponential & Trig Derivatives

Formula Booklet
$$(e^x)' = e^x, \quad (\sin x)' = \cos x, \quad (\cos x)' = -\sin x$$

Standard calculus derivatives listed in the IB AA formula booklet Topic 5.

Critical IB Examination Pitfalls

  • Reversing the Quotient Rule subtraction: The rule is $u'v - uv'$, NOT $uv' - u'v$. Writing $uv' - u'v$ introduces an overall negative sign error across all marks.
  • Forgetting to differentiate the inner function in Chain Rule: $\frac{d}{dx}(\sin(3x)) = 3\cos(3x)$, NOT $\cos(3x)$. Multiply by the inner derivative.
  • Using tangent slope for normal line: The normal gradient is the negative reciprocal $m_N = -\frac{1}{m_T}$. If $m_T = 4$, $m_N = -\frac{1}{4}$.
  • Expanding $(v)^2$ unnecessarily: Leaving the denominator as $(x^2 + 1)^2$ is preferred and prevents algebraic errors.

Section A: Diagnostic Concept Check

Instant feedback questions to test product, quotient, chain, and normal line rules.

Score: 0 / 4

1. What is the derivative of $f(x) = (2x - 3)^5$ with respect to $x$?

Correct! By the Chain Rule, $\frac{d}{dx}[(2x - 3)^5] = 5(2x - 3)^4 \cdot \frac{d}{dx}(2x - 3) = 5(2x - 3)^4(2) = 10(2x - 3)^4$.

2. If $y = x e^x$, what is $\frac{dy}{dx}$ at $x = 1$?

Correct! By the Product Rule, $\frac{dy}{dx} = (1)e^x + x e^x = (1 + x)e^x$. At $x = 1$, $(1 + 1)e^1 = 2e$.

3. The tangent to a curve at point $P(2, 5)$ has gradient $m_T = -\frac{1}{4}$. What is the equation of the normal line at $P$?

Correct! The normal gradient is $m_N = -\frac{1}{-1/4} = 4$. Using point-slope form with $P(2, 5)$: $y - 5 = 4(x - 2)$.

4. By the Quotient Rule, $\frac{d}{dx}\left(\frac{\sin x}{x}\right) = $

Correct! $u = \sin x \implies u' = \cos x$; $v = x \implies v' = 1$. $\frac{u'v - uv'}{v^2} = \frac{(\cos x)(x) - (\sin x)(1)}{x^2} = \frac{x\cos x - \sin x}{x^2}$.

Section B: Examination Practice

Multi-step questions with markscheme breakdowns.

Paper 1 (Non-Calculator) [6 marks]

Let $f(x) = \frac{e^x}{2x + 1}$ for $x \ne -0.5$.

(a) Show that $f'(x) = \frac{e^x(2x - 1)}{(2x + 1)^2}$. [3 marks]
(b) Find the equation of the normal line to the graph of $f$ at the point where $x = 0$. [3 marks]
View Markscheme
(a)
M1 for Quotient Rule with correct numerator subtraction order:
$$f'(x) = \frac{(e^x)(2x + 1) - (e^x)(2)}{(2x + 1)^2}$$ M1 for factoring $e^x$:
$$f'(x) = \frac{e^x(2x + 1 - 2)}{(2x + 1)^2}$$ $$f'(x) = \frac{e^x(2x - 1)}{(2x + 1)^2} \quad \text{[A1]}$$
(b)
M1 for finding $f(0)$: $f(0) = \frac{e^0}{2(0) + 1} = 1 \implies (0, 1)$
$f'(0) = \frac{e^0(2(0) - 1)}{(2(0) + 1)^2} = \frac{1(-1)}{1^2} = -1$
M1 for normal gradient: $m_N = -\frac{1}{-1} = 1$
Equation: $y - 1 = 1(x - 0) \implies \mathbf{y = x + 1}$ A1
Paper 2 (Calculator Allowed) [6 marks]

A curve is given by $y = (x^2 - 3)\ln(x)$ for $x > 0$.

(a) Find $\frac{dy}{dx}$. [2 marks]
(b) Find the equation of the tangent line to the curve at $x = 1$. [2 marks]
(c) Find the coordinates of the point on the curve where the tangent line is horizontal. Give your answer to 3 significant figures. [2 marks]
View Markscheme
(a)
M1 for Product Rule:
$$\frac{dy}{dx} = (2x)\ln(x) + (x^2 - 3)\left(\frac{1}{x}\right) = 2x\ln(x) + x - \frac{3}{x} \quad \text{[A1]}$$
(b)
At $x = 1$: $y = (1 - 3)\ln(1) = (-2)(0) = 0 \implies (1, 0)$
$m_T = 2(1)\ln(1) + 1 - \frac{3}{1} = 0 + 1 - 3 = -2$
$y - 0 = -2(x - 1) \implies \mathbf{y = -2x + 2}$ A1

(c)
M1 for setting $\frac{dy}{dx} = 0$: $2x\ln(x) + x - \frac{3}{x} = 0$
Using GDC numerical solver: $x \approx 1.34$ A1
$y = (1.336^2 - 3)\ln(1.336) \approx -0.352 \implies \mathbf{(1.34, -0.352)}$

Calculus Unit

Next Lessons in Calculus

5.1, 5.3 Limits and First Derivatives Rates of change, first principles limits, and power rule for polynomials. 5.2, 5.7, 5.8 Increasing, Decreasing & Stationary Points Monotonicity tests, second derivatives, local extrema, and inflection points. 5.4, 5.6 Tangents, Normals & Rules Product, quotient, and chain rules with tangent and normal line equations. 5.9 Kinematics with Derivatives Motion in a line: displacement, velocity, acceleration, and total distance. 5.5, 5.10, 5.11 Integration & Area Anti-differentiation, definite integrals, and area enclosed between curves.