Lecture Note 3: Differentiation

1. Meaning of Differentiation

Differentiation is the mathematical process of finding the rate at which one quantity changes with respect to another. If y = f(x), the derivative of y with respect to x measures how rapidly y changes when x changes.

Geometrically, the derivative of a function at a point is the gradient of the tangent line to the curve at that point. If the derivative is positive, the curve is increasing. If it is negative, the curve is decreasing. If it is zero, the curve may have a stationary point.

The derivative of f(x) may be written in several equivalent forms:

f′(x), ^dy⁄_dx, ^d⁄_dx[f(x)], or D_xy.

Key Takeaways

• Differentiation finds rate of change.
• The derivative at a point is the gradient of the tangent at that point.
• If y = f(x), then ^dy⁄_dx means the derivative of y with respect to x.
• Differentiation is the foundation for curve sketching, optimization, motion, and many applications in science and engineering.

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2. Differentiation from First Principles

The derivative of a function may be obtained directly from the limit definition. This is called differentiation from first principles.

For a function y = f(x), the average gradient between x and x + h is:

^{f(x + h) - f(x)}⁄_h

As h approaches zero, the second point approaches the first point, and the secant line becomes a tangent line. Therefore, the derivative is:

f′(x) = lim_{h → 0} ^{f(x + h) - f(x)}⁄_h

This formula is the fundamental definition of the derivative.

Example 1: Derivative of f(x) = x

Let f(x) = x. Then:

f(x + h) = x + h

Using first principles:

f′(x) = lim_{h → 0} ^{(x + h) - x}⁄_h

= lim_{h → 0} ^h⁄_h

= 1

Therefore:

^d⁄_dx(x) = 1

Example 2: Derivative of f(x) = 2x²

Let f(x) = 2x². Then:

f(x + h) = 2(x + h)²

= 2(x² + 2xh + h²)

= 2x² + 4xh + 2h²

Hence:

f′(x) = lim_{h → 0} ^{2x2 + 4xh + 2h2 - 2x2}⁄_h

= lim_{h → 0} ^{4xh + 2h2}⁄_h

= lim_{h → 0} (4x + 2h)

= 4x

Therefore:

^d⁄_dx(2x²) = 4x

Example 3: Derivative of f(x) = ^x⁄_{x + 1}

Let:

f(x) = ^x⁄_{x + 1}

Then:

f(x + h) = ^{x + h}⁄_{x + h + 1}

Using first principles:

f′(x) = lim_{h → 0} ^{x + h⁄_{x + h + 1} - x⁄_{x + 1}}⁄_h

Combining the fractions in the numerator gives:

f′(x) = lim_{h → 0} ^{(x + h)(x + 1) - x(x + h + 1)}⁄_{h(x + h + 1)(x + 1)}

Expanding the numerator:

(x + h)(x + 1) - x(x + h + 1)

= x² + xh + x + h - x² - xh - x

= h

Therefore:

f′(x) = lim_{h → 0} ^h⁄_{h(x + h + 1)(x + 1)}

= lim_{h → 0} ¹⁄_{(x + h + 1)(x + 1)}

= ¹⁄_{(x + 1)²}

Exam Points

• First principles always begins with f′(x) = lim_{h → 0} ^{f(x + h) - f(x)}⁄_h.
• Substitute x + h carefully wherever x appears.
• Simplify until the factor h cancels.
• Only after cancellation should h be allowed to approach zero.

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3. Standard Differentiation Rules

First principles is fundamental, but it is often too long for routine work. Differentiation rules allow us to find derivatives quickly and accurately.

3.1 Derivative of a Constant

If c is a constant, then:

^d⁄_dx(c) = 0

Example:

If y = 7, then ^dy⁄_dx = 0.

3.2 Power Rule

If y = xⁿ, where n is a real number, then:

^d⁄_dx(xⁿ) = nx^{n - 1}

Examples

1. If y = x³, then:

^dy⁄_dx = 3x²

2. If y = √x, rewrite it as y = x^1/2. Then:

^dy⁄_dx = ¹⁄₂x^-1/2

= ¹⁄_2√x

3. If y = ⁴⁄_x², rewrite as y = 4x^-2. Then:

^dy⁄_dx = 4(-2)x^-3

= -⁸⁄_x³

3.3 Constant Multiple Rule

If c is constant and f is differentiable, then:

^d⁄_dx[cf(x)] = c f′(x)

Example:

If y = 6x⁴, then:

^dy⁄_dx = 24x³

3.4 Sum and Difference Rules

If f and g are differentiable, then:

^d⁄_dx[f(x) + g(x)] = f′(x) + g′(x)

^d⁄_dx[f(x) - g(x)] = f′(x) - g′(x)

Example

Differentiate:

f(x) = -2x³ + 9x² - 2

Solution:

f′(x) = -6x² + 18x

Key Takeaways

• The derivative of a constant is zero.
• Use the power rule after writing radicals and reciprocals in index form.
• Differentiate sums and differences term by term.
• Always simplify negative powers where necessary.

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4. Product Rule

The product rule is used when two differentiable functions are multiplied.

If y = f(x)g(x), then:

^dy⁄_dx = f(x)g′(x) + g(x)f′(x)

In words: derivative of the first times the second, plus derivative of the second times the first.

Example

Differentiate:

h(x) = (3x - 2x²)(5 + 4x)

Solution:

Let u = 3x - 2x² and v = 5 + 4x.

Then:

u′ = 3 - 4x

v′ = 4

By the product rule:

h′(x) = u v′ + v u′

= (3x - 2x²)(4) + (5 + 4x)(3 - 4x)

= 12x - 8x² + 15 - 8x - 16x²

= -24x² + 4x + 15

Exam Points

• Use the product rule when two variable expressions are multiplied.
• Do not simply multiply the derivatives.
• You may expand first if the expression is simple, but the product rule is safer for longer products.

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5. Quotient Rule

The quotient rule is used when one differentiable function is divided by another.

If:

y = ^f(x)⁄_g(x)

then:

^dy⁄_dx = ^{g(x)f′(x) - f(x)g′(x)}⁄_[g(x)]²

Example

Differentiate:

y = ^{x2 + x - 2}⁄_{x³ + 6}

Solution:

Let f(x) = x² + x - 2 and g(x) = x³ + 6.

Then:

f′(x) = 2x + 1

g′(x) = 3x²

By the quotient rule:

^dy⁄_dx = ^{(x3 + 6)(2x + 1) - (x2 + x - 2)(3x2)}⁄_{(x³ + 6)²}

Expanding the numerator:

(x³ + 6)(2x + 1) = 2x⁴ + x³ + 12x + 6

(x² + x - 2)(3x²) = 3x⁴ + 3x³ - 6x²

Therefore:

^dy⁄_dx = ^{-x4 - 2x3 + 6x2 + 12x + 6}⁄_{(x³ + 6)²}

Exam Points

• Use the quotient rule when both numerator and denominator contain variable expressions.
• Remember the order: denominator times derivative of numerator minus numerator times derivative of denominator.
• The denominator is always squared.
• Some quotients are easier if rewritten with negative powers before differentiating.

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6. Chain Rule

The chain rule is used to differentiate composite functions. A composite function is a function inside another function.

If y = f(u) and u = g(x), then:

^dy⁄_dx = ^dy⁄_du × ^du⁄_dx

Example 1

Differentiate:

y = √(x² + 1)

Solution:

Let:

u = x² + 1

Then:

y = √u = u^1/2

^dy⁄_du = ¹⁄_2√u

^du⁄_dx = 2x

Therefore:

^dy⁄_dx = ¹⁄_2√u × 2x

= ^x⁄_√u

= ^x⁄_{√(x² + 1)}

Example 2

Differentiate:

y = (3x² - 4)⁴

Solution:

Let:

u = 3x² - 4

Then:

y = u⁴

^dy⁄_du = 4u³

^du⁄_dx = 6x

Therefore:

^dy⁄_dx = 4u³ × 6x

= 24x(3x² - 4)³

Example 3

Differentiate:

f(x) = ¹⁄_{(3x - 4)²}

Solution:

Rewrite as:

f(x) = (3x - 4)^-2

Let:

u = 3x - 4

Then:

f(u) = u^-2

f′(u) = -2u^-3

u′(x) = 3

Therefore:

f′(x) = -2u^-3 × 3

= -6u^-3

= -⁶⁄_{(3x - 4)³}

Key Takeaways

• The chain rule is essential for powers of brackets, roots of functions, and trigonometric functions of functions.
• Differentiate the outer function first, leaving the inner function unchanged.
• Then multiply by the derivative of the inner function.
• When possible, rewrite roots and quotients using indices before differentiating.

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7. Derivatives of Trigonometric Functions

The standard trigonometric derivatives must be used with angles measured in radians.

Important Trigonometric Limits

lim_{x → 0} ^{sin x}⁄_x = 1

lim_{x → 0} ^{1 - cos x}⁄_x = 0

These results are used to prove the derivatives of sine and cosine.

Basic Trigonometric Derivatives

Function	Derivative
sin x	cos x
cos x	-sin x
tan x	sec2 x
cot x	-csc2 x
sec x	sec x tan x
csc x	-csc x cot x

Example 1

Differentiate:

y = sin(3x)

Solution:

Let u = 3x. Then y = sin u.

^dy⁄_dx = cos u × 3

= 3 cos(3x)

Example 2

Differentiate:

g(x) = sin(4x² - 3)

Solution:

Let u = 4x² - 3. Then:

g′(x) = cos u × 8x

= 8x cos(4x² - 3)

Example 3

Differentiate:

y = x⁶ cos x + 2x⁵

Solution:

Use the product rule on x⁶ cos x:

^dy⁄_dx = x⁶(-sin x) + cos x(6x⁵) + 10x⁴

= -x⁶ sin x + 6x⁵ cos x + 10x⁴

= x⁴(-x² sin x + 6x cos x + 10)

Example 4

Differentiate:

y = tan x

Solution:

Since tan x = ^{sin x}⁄_{cos x}, use the quotient rule:

^d⁄_dx(tan x) = ^{cos x(cos x) - sin x(-sin x)}⁄_{cos² x}

= ^{cos2 x + sin2 x}⁄_{cos² x}

= ¹⁄_{cos² x}

= sec² x

Exam Points

• Trigonometric differentiation assumes radians.
• The derivative of sin x is cos x.
• The derivative of cos x is -sin x.
• Use the chain rule when the angle is not simply x.
• Remember the negative signs in ^d⁄_dx(cos x), ^d⁄_dx(cot x), and ^d⁄_dx(csc x).

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8. Derivatives of Inverse Trigonometric Functions

Inverse trigonometric functions include sin^-1x, cos^-1x, and tan^-1x. These are also written as arcsin x, arccos x, and arctan x.

Standard Inverse Trigonometric Derivatives

Function	Derivative
sin-1x	1⁄√(1 - x2)
cos-1x	-1⁄√(1 - x2)
tan-1x	1⁄1 + x2

Example 1

Differentiate:

y = sin^-1(3x)

Solution:

Let u = 3x. Then:

^dy⁄_dx = ¹⁄_{√(1 - u²)} × 3

= ³⁄_{√(1 - 9x²)}

Example 2

Differentiate:

y = cos^-1(x²)

Solution:

Let u = x². Then:

^dy⁄_dx = -¹⁄_{√(1 - u²)} × 2x

= -^2x⁄_{√(1 - x⁴)}

Example 3

Differentiate:

y = tan^-1(²⁄_x)

Solution:

Let u = ²⁄_x = 2x^-1. Then:

^du⁄_dx = -2x^-2 = -²⁄_x²

Since ^d⁄_du(tan^-1u) = ¹⁄_{1 + u²},

^dy⁄_dx = ¹⁄_{1 + (²⁄x)²} × (-²⁄_x²)

= -²⁄_{x² + 4}

Key Takeaways

• Use the chain rule for inverse trigonometric functions of another expression.
• The derivative of cos^-1x has a negative sign.
• The expression under the square root in the derivatives of sin^-1x and cos^-1x is 1 - x².
• Simplify carefully when the inner function is a fraction.

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9. Implicit Differentiation

A function is given explicitly when y is written directly in terms of x, such as y = x² + 1.

A function is given implicitly when x and y are mixed together in one equation, such as:

x²y - 5x = 3

or:

x³ + y³ = 3xy

When differentiating implicitly, treat y as a function of x. Therefore:

^d⁄_dx(y) = ^dy⁄_dx

^d⁄_dx(yⁿ) = ny^{n - 1dy}⁄_dx

Example 1

Find ^dy⁄_dx if:

x²y - 5x = 3

Solution:

Differentiate term by term:

^d⁄_dx(x²y) - ^d⁄_dx(5x) = ^d⁄_dx(3)

Using the product rule on x²y:

x^2dy⁄_dx + 2xy - 5 = 0

Therefore:

x^2dy⁄_dx = 5 - 2xy

^dy⁄_dx = ^{5 - 2xy}⁄_x²

Example 2

Find ^dy⁄_dx if:

x³ + y³ = 3xy

Solution:

Differentiate both sides:

3x² + 3y^2dy⁄_dx = 3(x^dy⁄_dx + y)

Expand the right side:

3x² + 3y^2dy⁄_dx = 3x^dy⁄_dx + 3y

Collect terms containing ^dy⁄_dx:

3y^2dy⁄_dx - 3x^dy⁄_dx = 3y - 3x²

Factor:

(3y² - 3x)^dy⁄_dx = 3y - 3x²

Therefore:

^dy⁄_dx = ^{y - x2}⁄_{y² - x}

Example 3

Find ^dy⁄_dx if:

x² + y² + sin y = 3

Solution:

Differentiate term by term:

2x + 2y^dy⁄_dx + cos y ^dy⁄_dx = 0

Factor ^dy⁄_dx:

(2y + cos y)^dy⁄_dx = -2x

Therefore:

^dy⁄_dx = -^2x⁄_{2y + cos y}

Exam Points

• Implicit differentiation is used when y is not isolated.
• Whenever you differentiate a term involving y, multiply by ^dy⁄_dx.
• Use the product rule when x and y are multiplied.
• After differentiating, collect all ^dy⁄_dx terms on one side and factor.

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10. Higher Derivatives

If y = f(x), then the first derivative is f′(x) or ^dy⁄_dx. Since the derivative is often another function of x, it can also be differentiated.

The derivative of the first derivative is called the second derivative and is written as:

^d2y⁄_dx² or f″(x)

The third derivative is written as:

^d3y⁄_dx³

In general, the nth derivative is written as:

^dny⁄_dxⁿ

Example

If:

y = ^{cos x}⁄_{1 - sin x}

find ^d2y⁄_dx².

Solution:

Using the quotient rule:

^dy⁄_dx = ^{(1 - sin x)(-sin x) - cos x(-cos x)}⁄_{(1 - sin x)²}

= ^{-sin x + sin2x + cos2x}⁄_{(1 - sin x)²}

= ^{1 - sin x}⁄_{(1 - sin x)²}

= ¹⁄_{1 - sin x}

Differentiate again:

^d2y⁄_dx² = ^d⁄_dx[(1 - sin x)^-1]

= -(1 - sin x)^-2(-cos x)

= ^{cos x}⁄_{(1 - sin x)²}

Key Takeaways

• The second derivative is the derivative of the first derivative.
• Higher derivatives are found by repeated differentiation.
• The second derivative is useful for testing maximum and minimum values.
• Use clear notation so that ^d2y⁄_dx² is not confused with (^dy⁄_dx)².

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11. Derivatives of Logarithmic Functions

The number e is a special mathematical constant defined by:

e = lim_{n → ∞} (1 + ¹⁄_n)ⁿ

Its approximate value is:

e ≈ 2.71828

The function e^x is called the natural exponential function. Its inverse is the natural logarithm, written as:

ln x

Important identities:

ln(e^x) = x

e^{ln x} = x

Logarithmic Derivative Rules

^d⁄_dx(ln x) = ¹⁄_x

If u is a differentiable function of x, then:

^d⁄_dx(ln u) = ¹⁄_u × ^du⁄_dx

Example 1

Differentiate:

y = ln(x³ + 1)

Solution:

Let u = x³ + 1. Then:

^dy⁄_dx = ¹⁄_u × ^du⁄_dx

= ¹⁄_{x³ + 1} × 3x²

= ^3x2⁄_{x³ + 1}

Example 2

Differentiate:

y = ln(sec x)

Solution:

^dy⁄_dx = ¹⁄_{sec x} × sec x tan x

= tan x

Example 3

Differentiate:

y = ln(^{x + 1}⁄_{√(x - 2)})

Solution:

Use logarithmic laws first:

y = ln(x + 1) - ln(√(x - 2))

Since √(x - 2) = (x - 2)^1/2,

y = ln(x + 1) - ¹⁄₂ ln(x - 2)

Therefore:

^dy⁄_dx = ¹⁄_{x + 1} - ¹⁄_{2(x - 2)}

= ^{x - 5}⁄_{2(x + 1)(x - 2)}

Logarithms to Other Bases

If a > 0 and a ≠ 1, then:

log_ax = ^{ln x}⁄_{ln a}

Therefore:

^d⁄_dx(log_ax) = ¹⁄_{x ln a}

Example

Differentiate:

f(x) = log₁₀(2 + sin x)

Solution:

f′(x) = ^{cos x}⁄_{(2 + sin x)ln 10}

Exam Points

• The derivative of ln x is ¹⁄_x.
• For ln u, multiply by u′.
• Use logarithmic laws to simplify products, quotients, and powers before differentiating.
• For logarithms not in base e, convert using log_ax = ^{ln x}⁄_{ln a}.

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12. Derivatives of Exponential Functions

Natural Exponential Function

The derivative of the natural exponential function is itself:

^d⁄_dx(e^x) = e^x

If u is a differentiable function of x, then:

^d⁄_dx(e^u) = e^udu⁄_dx

Example 1

Differentiate:

y = e^{2x2 - 1}

Solution:

Let u = 2x² - 1. Then:

^du⁄_dx = 4x

Therefore:

^dy⁄_dx = 4x e^{2x2 - 1}

Example 2

Differentiate:

y = e^{tan x}

Solution:

^dy⁄_dx = e^{tan x} sec²x

Example 3

Differentiate:

y = x²e^-3/x

Solution:

Use the product rule:

^dy⁄_dx = x^2d⁄_dx(e^-3/x) + e^-3/xd⁄_dx(x²)

Since ^d⁄_dx(-³⁄_x) = ³⁄_x²,

^dy⁄_dx = x²e^-3/x(³⁄_x²) + 2xe^-3/x

= 3e^-3/x + 2xe^-3/x

= (3 + 2x)e^-3/x

Exponential Functions with Other Bases

If a > 0 and a ≠ 1, then:

^d⁄_dx(a^x) = a^xln a

If u is a differentiable function of x, then:

^d⁄_dx(a^u) = a^uln a × ^du⁄_dx

Example

Differentiate:

y = 10^x2

Solution:

Let u = x². Then:

^dy⁄_dx = 10^uln 10 × 2x

= 2x(ln 10)10^x2

Key Takeaways

• The derivative of e^x is e^x.
• For e^u, multiply by u′.
• For a^x, multiply by ln a.
• Use product rule when exponentials multiply other functions of x.

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13. Parametric Differentiation

Sometimes x and y are both expressed in terms of a third variable, usually t. Such equations are called parametric equations.

If:

x = x(t) and y = y(t)

then:

^dy⁄_dx = ^dy⁄_dt⁄_^dx⁄dt

provided ^dx⁄_dt ≠ 0.

Example 1

Find ^dy⁄_dx if:

x = t³ + t and y = 2t²

Solution:

^dy⁄_dt = 4t

^dx⁄_dt = 3t² + 1

Therefore:

^dy⁄_dx = ^4t⁄_{3t² + 1}

Example 2

Find ^dy⁄_dx if:

x = sin(2t) and y = cos(4t)

Solution:

^dy⁄_dt = -4sin(4t)

^dx⁄_dt = 2cos(2t)

Therefore:

^dy⁄_dx = ^-4sin(4t)⁄_2cos(2t) = -^2sin(4t)⁄_cos(2t)

Exam Points

• Differentiate x and y separately with respect to the parameter.
• Divide ^dy⁄_dt by ^dx⁄_dt.
• Do not attempt to treat y directly as a function of x unless the parameter is first eliminated.

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14. Stationary Points, Maximum Values, and Minimum Values

A stationary point occurs at a point on a curve where:

^dy⁄_dx = 0

At such points, the tangent to the curve is horizontal.

Stationary points may be classified as:

• Local maximum points.
• Local minimum points.
• Stationary points of inflection.

Second Derivative Test

Suppose ^dy⁄_dx = 0 at x = a.

• If ^d2y⁄_dx² < 0 at x = a, the point is a local maximum.
• If ^d2y⁄_dx² > 0 at x = a, the point is a local minimum.
• If ^d2y⁄_dx² = 0, the test is inconclusive; further analysis is required.

Example

Let:

y = x³ - 2x² + x + 4

Find the stationary points and determine whether they give maximum or minimum values.

Solution:

First derivative:

^dy⁄_dx = 3x² - 4x + 1

At stationary points:

3x² - 4x + 1 = 0

Factor:

(3x - 1)(x - 1) = 0

Thus:

x = ¹⁄₃ or x = 1

Second derivative:

^d2y⁄_dx² = 6x - 4

At x = ¹⁄₃:

6(¹⁄₃) - 4 = 2 - 4 = -2

Since this is negative, x = ¹⁄₃ gives a local maximum.

The maximum value is:

y = (¹⁄₃)³ - 2(¹⁄₃)² + ¹⁄₃ + 4

= ¹⁄₂₇ - ²⁄₉ + ¹⁄₃ + 4

= ¹¹²⁄₂₇

At x = 1:

6(1) - 4 = 2

Since this is positive, x = 1 gives a local minimum.

The minimum value is:

y = 1³ - 2(1)² + 1 + 4 = 4

Key Takeaways

• Stationary points occur where the first derivative is zero.
• Use the second derivative to classify stationary points.
• A negative second derivative indicates a local maximum.
• A positive second derivative indicates a local minimum.
• If the second derivative is zero, use another test because the result is inconclusive.

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15. Exam-Focused Formula Summary

First Principle

f′(x) = lim_{h → 0} ^{f(x + h) - f(x)}⁄_h

Basic Rules

^d⁄_dx(c) = 0

^d⁄_dx(xⁿ) = nx^{n - 1}

^d⁄_dx[cf(x)] = cf′(x)

^d⁄_dx[f(x) ± g(x)] = f′(x) ± g′(x)

Product and Quotient Rules

^d⁄_dx[f(x)g(x)] = f(x)g′(x) + g(x)f′(x)

^d⁄_dx[^f(x)⁄_g(x)] = ^{g(x)f′(x) - f(x)g′(x)}⁄_[g(x)]²

Chain Rule

^dy⁄_dx = ^dy⁄_du × ^du⁄_dx

Trigonometric Derivatives

^d⁄_dx(sin x) = cos x

^d⁄_dx(cos x) = -sin x

^d⁄_dx(tan x) = sec²x

^d⁄_dx(cot x) = -csc²x

^d⁄_dx(sec x) = sec x tan x

^d⁄_dx(csc x) = -csc x cot x

Inverse Trigonometric Derivatives

^d⁄_dx(sin^-1x) = ¹⁄_{√(1 - x²)}

^d⁄_dx(cos^-1x) = -¹⁄_{√(1 - x²)}

^d⁄_dx(tan^-1x) = ¹⁄_{1 + x²}

Logarithmic and Exponential Derivatives

^d⁄_dx(ln x) = ¹⁄_x

^d⁄_dx(ln u) = ^u′⁄_u

^d⁄_dx(e^x) = e^x

^d⁄_dx(e^u) = e^uu′

^d⁄_dx(a^x) = a^xln a

^d⁄_dx(a^u) = a^uln a × u′

Parametric Differentiation

^dy⁄_dx = ^dy⁄_dt⁄_^dx⁄dt

Stationary Points

^dy⁄_dx = 0

Maximum if ^d2y⁄_dx² < 0.

Minimum if ^d2y⁄_dx² > 0.

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16. Practice Questions for Revision

A. First Principles

1. Find the derivative of y = x³ + 2x from first principles.
2. Find the derivative of f(x) = x² + x from first principles.
3. Find the derivative of g(x) = 4x² + 2 from first principles.
4. Find the derivative of y = x⁴ - x² from first principles.

B. Basic Differentiation Rules

1. Differentiate y = 7x⁵ - 3x⁴ + x².
2. Differentiate y = 2t⁴ - 6t + t - ²⁄_t².
3. Differentiate y = 6³√x.

C. Product and Quotient Rules

1. Differentiate f(x) = (x³ + 1)(x⁴ + 1).
2. Differentiate g(x) = (x + x³)(x² - 1).
3. Find ^dy⁄_dx if y = ^x2⁄_{x + 1}.
4. Find ^dy⁄_dx if y = ^x2⁄_{1 + √x}.

D. Chain Rule and Trigonometric Functions

1. Differentiate y = ²⁄_{(x - 13)⁵}.
2. Differentiate f(x) = (^{sin x}⁄_{1 + cos x})³.
3. Differentiate g(x) = (x - ²⁄_x)².
4. Differentiate h(x) = (3x² - 4)⁴.
5. Find the derivative of y = x²tan²(4x).

E. Inverse Trigonometric, Logarithmic, and Exponential Functions

1. Differentiate y = sin^-1(√(1 - x²)).
2. Differentiate y = cos^-1(^5x⁄₂).
3. Differentiate y = tan^-1(^x⁄_a).
4. Differentiate y = e^{sin x}.
5. Differentiate y = e^-2x2.
6. Differentiate f(x) = 2^x3.
7. Differentiate y = ln(√(x² - 2)).

F. Implicit and Parametric Differentiation

1. Differentiate implicitly: xy = 5.
2. Differentiate implicitly: x²y² + x + y = 0.
3. Differentiate implicitly: xy + sin y = 2.
4. Find ^dy⁄_dx if y = sin θ and x = 3cos θ.
5. Find ^dy⁄_dx if y = t² and x = ¹⁄_t.

G. Higher Derivatives and Stationary Points

1. Find ^d2y⁄_dx² if y = cos²x.
2. Find ^d2y⁄_dx² if y = ¹⁄_{2 - x}.
3. Find the stationary points and classify them for y = 2x³ + 3x² - 36x + 5.
4. Find the stationary points and classify them for y = x⁴ - 4x³.

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17. Final Revision Checklist

Before an examination, make sure you can:

• Define differentiation as a rate of change.
• Find derivatives from first principles.
• Apply the constant, power, sum, and difference rules.
• Use the product rule and quotient rule correctly.
• Use the chain rule for composite functions.
• Differentiate trigonometric and inverse trigonometric functions.
• Differentiate logarithmic and exponential functions.
• Differentiate implicit functions.
• Differentiate parametric equations.
• Find second and higher derivatives.
• Find and classify stationary points using the second derivative test.