Strand 5 — Functions & Calculus

Differentiation

5th Year · 6th Year (Leaving Cert)

✓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 find the derivatives of trigonometric, exponential, and logarithmic functions (HL).
✓By the end of this lesson students will be able to solve problems involving tangents, normals, maxima, minima, and rates of change.
✓By the end of this lesson students will be able to use the second derivative to determine concavity and points of inflection (HL).

Key concepts

Differentiation from First Principles

Differentiation from first principles is the fundamental definition of the derivative of a function. It involves finding the limit of the average rate of change as the interval approaches zero. This method helps us understand where the differentiation rules come from.

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 functions, say u(x) and v(x).

If y = u v, then dy/dx = 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 functions, say u(x) and v(x).

If y = u / v, then dy/dx = [v (du/dx) - u (dv/dx)] / v²

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 y can be differentiated with respect to x.

If y = f(u) and u = g(x), then dy/dx = (dy/du) * (du/dx)

Derivatives of Trigonometric Functions

These are standard derivatives for trigonometric functions.

d/dx (sin x) = cos x d/dx (cos x) = -sin x d/dx (tan x) = sec² x

Derivatives of Exponential and Logarithmic Functions (HL)

These are standard derivatives for exponential and logarithmic functions, particularly important for Higher Level students.

d/dx (e^x) = e^x d/dx (ln x) = 1/x d/dx (a^x) = a^x ln a d/dx (log_a x) = 1/(x ln a)

Tangents and Normals

The derivative f'(x) gives the slope (gradient) of the tangent to the curve y = f(x) at any point (x, y). The normal to the curve at a point is a line perpendicular to the tangent at that point.

Slope of tangent (m_t) = f'(x) Slope of normal (m_n) = -1 / m_t (if m_t ≠ 0)

Maxima and Minima (Optimisation)

Local maxima and minima (turning points or stationary points) occur where the first derivative of a function is zero, i.e., f'(x) = 0. The nature of these points can be determined using the first derivative test (sign change of f'(x)) or the second derivative test (sign of f''(x)). Optimisation problems involve finding the maximum or minimum value of a quantity.

For a stationary point, f'(x) = 0. If f''(x) > 0, it's a local minimum. If f''(x) < 0, it's a local maximum. If f''(x) = 0, the test is inconclusive.

Rates of Change

Rates of change describe how one quantity changes with respect to another, often time. These problems frequently involve implicit differentiation and the chain rule.

If a quantity V changes with respect to time t, its rate of change is dV/dt.

Second Derivative and Points of Inflection (HL)

The second derivative, f''(x), describes the rate of change of the first derivative and indicates the concavity of the curve. A point of inflection is a point where the concavity of the curve changes, and f''(x) = 0 at such a point (though f''(x) = 0 does not guarantee an inflection point; the sign of f''(x) must change).

Concave up: f''(x) > 0 Concave down: f''(x) < 0 Point of inflection: f''(x) = 0 and f''(x) changes sign.

Key facts to remember

1The derivative f'(x) represents the instantaneous rate of change of f(x) with respect to x, and the slope of the tangent to the curve y = f(x).
2Differentiation from first principles uses the limit definition: f'(x) = lim (h→0) [f(x+h) - f(x)] / h.
3Key differentiation rules are the Product Rule (uv)' = u'v + uv', Quotient Rule (u/v)' = (u'v - uv')/v², and Chain Rule (f(g(x)))' = f'(g(x))g'(x).
4Standard derivatives include: d/dx (xⁿ) = nxⁿ⁻¹, d/dx (sin x) = cos x, d/dx (cos x) = -sin x, d/dx (tan x) = sec² x.
5(HL) Standard derivatives include: d/dx (e^x) = e^x, d/dx (ln x) = 1/x, d/dx (a^x) = a^x ln a, d/dx (log_a x) = 1/(x ln a).
6Stationary points (maxima, minima, points of inflection) occur where f'(x) = 0.
7The second derivative f''(x) determines concavity: f''(x) > 0 for concave up, f''(x) < 0 for concave down.
8(HL) A point of inflection occurs where f''(x) = 0 and the concavity changes (i.e., f''(x) changes sign).

Worked examples

Example 1

Given the function f(x) = (3x² - 1)⁵, find f'(x) and hence find the equation of the tangent to the curve at the point where x = 1.

IStep 1: Identify the function as a composite function and apply the chain rule.

IILet u = 3x² - 1, then f(x) = u⁵.

IIIdu/dx = d/dx (3x² - 1) = 6x.

IVdf/du = d/du (u⁵) = 5u⁴.

VStep 2: Apply the chain rule: df/dx = (df/du) * (du/dx).

VIf'(x) = 5u⁴ * 6x = 5(3x² - 1)⁴ * 6x = 30x(3x² - 1)⁴.

VIIStep 3: Find the slope of the tangent at x = 1.

VIIIf'(1) = 30(1)(3(1)² - 1)⁴ = 30(1)(3 - 1)⁴ = 30(2)⁴ = 30(16) = 480.

9So, the slope of the tangent, m = 480.

10Step 4: Find the y-coordinate of the point on the curve where x = 1.

11f(1) = (3(1)² - 1)⁵ = (3 - 1)⁵ = 2⁵ = 32.

12The point is (1, 32).

13Step 5: Use the point-slope form of a line (y - y₁) = m(x - x₁) to find the equation of the tangent.

14y - 32 = 480(x - 1)

15y - 32 = 480x - 480

16y = 480x - 448.

Answer

The equation of the tangent to the curve at x = 1 is y = 480x - 448.

Remember to substitute back the original expression for 'u' after differentiating with respect to 'u'.

Example 2

(HL) A cylindrical can with a lid is to be made to hold 1000 cm³ of liquid. Find the dimensions of the can that will minimise the amount of material used.

IStep 1: Define variables and set up equations for volume and surface area.

IILet r be the radius and h be the height of the cylinder.

IIIVolume V = πr²h = 1000 cm³.

IVSurface Area A = 2πr² (for top and bottom) + 2πrh (for curved surface).

VStep 2: Express the quantity to be minimised (Surface Area) in terms of a single variable.

VIFrom V = 1000, we have h = 1000 / (πr²).

VIISubstitute h into the surface area formula:

VIIIA = 2πr² + 2πr(1000 / (πr²))

9A = 2πr² + 2000/r.

10Step 3: Differentiate the surface area function with respect to r and set it to zero.

11A = 2πr² + 2000r⁻¹.

12dA/dr = 4πr - 2000r⁻² = 4πr - 2000/r².

13Set dA/dr = 0:

144πr - 2000/r² = 0

154πr = 2000/r²

164πr³ = 2000

17r³ = 2000 / (4π) = 500 / π.

18r = ³√(500/π) ≈ 5.419 cm (to 3 decimal places).

19Step 4: Use the second derivative test to confirm it's a minimum.

20d²A/dr² = d/dr (4πr - 2000r⁻²) = 4π + 4000r⁻³ = 4π + 4000/r³.

21Since r > 0, r³ > 0, so 4000/r³ > 0. Therefore, d²A/dr² > 0, which confirms it's a minimum.

22Step 5: Calculate the corresponding height h.

23h = 1000 / (πr²) = 1000 / (π * (³√(500/π))²) = 1000 / (π * (500/π)^(2/3)).

24Alternatively, notice that r³ = 500/π, so πr² = 500/r.

25h = 1000 / (500/r) = 2r.

26So, h = 2 * ³√(500/π) ≈ 2 * 5.419 = 10.838 cm.

27The height is twice the radius, meaning the diameter equals the height.

Answer

The dimensions that minimise the material used are radius r = ³√(500/π) cm (approx. 5.42 cm) and height h = 2 * ³√(500/π) cm (approx. 10.84 cm).

For optimisation problems, always verify the nature of the stationary point using the second derivative test or by checking the sign change of the first derivative.

Example 3

(HL) Find the coordinates of any points of inflection for the function f(x) = x⁴ - 4x³ + 10.

IStep 1: Find the first derivative, f'(x).

IIf'(x) = d/dx (x⁴ - 4x³ + 10) = 4x³ - 12x².

IIIStep 2: Find the second derivative, f''(x).

IVf''(x) = d/dx (4x³ - 12x²) = 12x² - 24x.

VStep 3: Set the second derivative to zero and solve for x to find potential points of inflection.

VI12x² - 24x = 0

VII12x(x - 2) = 0

VIIISo, x = 0 or x = 2.

9Step 4: Check the sign of f''(x) around these x-values to confirm if concavity changes.

10For x = 0:

11Test x = -1 (left of 0): f''(-1) = 12(-1)² - 24(-1) = 12 + 24 = 36 (> 0, concave up).

12Test x = 1 (right of 0): f''(1) = 12(1)² - 24(1) = 12 - 24 = -12 (< 0, concave down).

13Since the sign of f''(x) changes at x = 0, there is a point of inflection at x = 0.

14For x = 2:

15Test x = 1 (left of 2): f''(1) = -12 (< 0, concave down).

16Test x = 3 (right of 2): f''(3) = 12(3)² - 24(3) = 12(9) - 72 = 108 - 72 = 36 (> 0, concave up).

17Since the sign of f''(x) changes at x = 2, there is a point of inflection at x = 2.

18Step 5: Find the y-coordinates for these x-values using the original function f(x).

19For x = 0: f(0) = (0)⁴ - 4(0)³ + 10 = 10. Point: (0, 10).

20For x = 2: f(2) = (2)⁴ - 4(2)³ + 10 = 16 - 4(8) + 10 = 16 - 32 + 10 = -6. Point: (2, -6).

Answer

The points of inflection are (0, 10) and (2, -6).

Remember that f''(x) = 0 is a necessary condition for a point of inflection, but not sufficient. You must verify a change in the sign of the second derivative.

Common mistakes

✗Confusing the product rule with the quotient rule, or applying them incorrectly.
✗Forgetting to apply the chain rule when differentiating composite functions (e.g., (2x+3)⁴ or sin(3x)).
✗Incorrectly calculating the slope of the normal by not taking the negative reciprocal of the tangent's slope.
✗Not checking the nature of stationary points (maxima/minima) after finding f'(x) = 0, or incorrectly interpreting the second derivative test.
✗Failing to find the y-coordinate of points (e.g., for tangents, normals, or points of inflection) after finding the x-coordinate.

Exam tips

★Always show all steps in your differentiation, especially when using rules like the product, quotient, or chain rule. This helps in getting partial credit.
★For optimisation problems, clearly define your variables, set up the function to be optimised, differentiate, and verify the nature of the stationary point.
★When dealing with tangents and normals, remember that the slope of the tangent is f'(x) and the slope of the normal is -1/f'(x). Use the point-slope formula for the equation of a line.
★(HL) For points of inflection, ensure you test the sign of the second derivative on either side of the potential inflection point to confirm a change in concavity.

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