GELE Mathematics — Differential CalculusExam Answer Templates

Marks

1

Topic

Derivative Rules

Difficulty

easy

Template Id

T1

Examiner Tip

Write the formula in Leibniz notation d/dx — mixed notation loses the mark on strict marking schemes.

Model Answer

Power Rule: d/dx (x^n) = n·x^(n−1), where n is any real number.

Question Type

very_short_answer

Answer Structure

• Line 1: Write the formula d/dx(x^n) = n·x^(n-1) with correct notation [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing nx^n instead of nx^(n-1) — exponent not reduced
• Omitting the coefficient n in front of x
• Restricting n to integers only when the rule applies to all real n

Key Phrases To Include

• d/dx
• x^n
• n·x^(n−1)
• real number

Marks

2

Topic

Derivative Rules — Power Rule

Difficulty

easy

Template Id

T2

Examiner Tip

Write each term's derivative on a separate line — this makes it easier for the examiner to award partial marks if one term is wrong.

Model Answer

Given: y = 5x^4 − 3x^2 + 7x − 2 Required: y′ and y′(1) Solution: y′ = d/dx(5x^4) − d/dx(3x^2) + d/dx(7x) − d/dx(2) y′ = 20x^3 − 6x + 7 At x = 1: y′(1) = 20(1)^3 − 6(1) + 7 y′(1) = 20 − 6 + 7 ∴ y′(1) = 21

Question Type

numerical

Answer Structure

• Step 1: Apply power rule term-by-term to obtain y′ = 20x^3 − 6x + 7 [1 mark]
• Step 2: Substitute x = 1 and evaluate to get 21 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Forgetting the derivative of the constant −2 is zero (should simply vanish)
• Arithmetic error in evaluating 20(1) − 6(1) + 7
• Not writing the derivative expression before substituting

Key Phrases To Include

• y′ = 20x^3 − 6x + 7
• power rule
• term-by-term
• y′(1) = 21

Marks

2

Topic

Derivative Rules — Product Rule

Difficulty

easy

Template Id

T3

Examiner Tip

Explicitly label u, v, u′, v′ before writing the formula — this structured approach earns the method mark even if arithmetic fails.

Model Answer

Given: y = (3x^2 + 1)(2x − 5) Required: y′ Solution: Let u = 3x^2 + 1 → u′ = 6x Let v = 2x − 5 → v′ = 2 Product Rule: y′ = u′v + uv′ y′ = (6x)(2x − 5) + (3x^2 + 1)(2) y′ = 12x^2 − 30x + 6x^2 + 2 ∴ y′ = 18x^2 − 30x + 2

Question Type

numerical

Answer Structure

• Step 1: Identify u and v, then compute u′ and v′ [0.5 mark — setup]
• Step 2: Apply Product Rule formula y′ = u′v + uv′ and state it [0.5 mark]
• Step 3: Substitute and expand correctly [0.5 mark]
• Step 4: Simplify to final form 18x^2 − 30x + 2 [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using uv′ − u′v (confusing with Quotient Rule numerator order)
• Errors in expanding (3x^2 + 1)(2) — forgetting to distribute the 2
• Not simplifying like terms in the final answer

Key Phrases To Include

• Product Rule
• u′v + uv′
• u = 3x^2 + 1
• v = 2x − 5
• y′ = 18x^2 − 30x + 2

Marks

2

Topic

Derivative Rules — Chain Rule

Difficulty

medium

Template Id

T4

Examiner Tip

Always write 'Chain Rule: dy/dx = f′(u)·u′' as a header — it signals to the examiner you know the method and earns the method mark.

Model Answer

Given: y = (4x^3 − 2x)^5 Required: y′ Solution: Let u = 4x^3 − 2x → u′ = 12x^2 − 2 Chain Rule: y′ = 5u^4 · u′ y′ = 5(4x^3 − 2x)^4 · (12x^2 − 2) ∴ y′ = 5(12x^2 − 2)(4x^3 − 2x)^4

Question Type

numerical

Answer Structure

• Step 1: Identify inner function u = 4x^3 − 2x and differentiate to get u′ = 12x^2 − 2 [1 mark]
• Step 2: Apply Chain Rule: bring down exponent 5, reduce power by 1, multiply by u′ [1 mark]

Scoring Breakdown

Common Mark Deductions

• Forgetting to multiply by the inner derivative u′ — the most common chain rule error
• Reducing the exponent to 4 but keeping original exponent in u — power reduction omitted
• Differentiating only the outer function without touching the inner

Key Phrases To Include

• Chain Rule
• inner function
• u′ = 12x^2 − 2
• 5(4x^3 − 2x)^4

Marks

2

Topic

Limits and Continuity

Difficulty

easy

Template Id

T5

Examiner Tip

Always attempt direct substitution first and explicitly state the result — 'Direct substitution gives 0/0 — indeterminate form' is a required step that earns a mark.

Model Answer

Given: lim(x→2) [(x^2 − 4)/(x − 2)] Solution: Direct substitution: (4 − 4)/(2 − 2) = 0/0 → Indeterminate form Method 1 — Factoring: (x^2 − 4)/(x − 2) = (x + 2)(x − 2)/(x − 2) = x + 2 (for x ≠ 2) lim(x→2) (x + 2) = 2 + 2 = 4 ∴ lim(x→2) [(x^2 − 4)/(x − 2)] = 4

Question Type

numerical

Answer Structure

• Step 1: Attempt direct substitution and identify 0/0 indeterminate form [0.5 mark]
• Step 2: Factor numerator as (x+2)(x−2) [0.5 mark]
• Step 3: Cancel (x−2) factor [0.5 mark]
• Step 4: Evaluate limit = 4 [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Concluding the limit does not exist because direct substitution gave 0/0 — must resolve the indeterminate form
• Not writing 'for x ≠ 2' after cancellation
• Arithmetic error after cancellation

Key Phrases To Include

• indeterminate form 0/0
• factoring
• (x+2)(x−2)
• cancel
• limit = 4

Marks

3

Topic

Limits and Continuity — L'Hôpital's Rule

Difficulty

medium

Template Id

T6

Examiner Tip

State 'L'Hôpital's Rule: since the form is 0/0, lim f/g = lim f′/g′' as a labelled step — this phrase alone earns the method mark.

Model Answer

Given: lim(x→0) [sin(3x)/x] Solution: Step 1 — Check indeterminate form: Direct substitution: sin(0)/0 = 0/0 → Indeterminate form ✓ L'Hôpital's Rule applies. Step 2 — Differentiate numerator and denominator separately: d/dx [sin(3x)] = 3cos(3x) d/dx [x] = 1 Step 3 — Apply L'Hôpital's Rule: lim(x→0) [sin(3x)/x] = lim(x→0) [3cos(3x)/1] Step 4 — Evaluate: = 3cos(3·0) = 3cos(0) = 3(1) ∴ lim(x→0) [sin(3x)/x] = 3

Question Type

numerical

Answer Structure

• Step 1: Verify 0/0 indeterminate form by direct substitution [1 mark]
• Step 2: Differentiate numerator sin(3x) → 3cos(3x) using chain rule [1 mark]
• Step 3: Form new limit and evaluate to get 3 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Applying L'Hôpital without first verifying the indeterminate form
• Differentiating sin(3x)/x as a quotient instead of differentiating numerator and denominator separately
• Forgetting the chain rule: writing d/dx[sin(3x)] = cos(3x) instead of 3cos(3x)

Key Phrases To Include

• indeterminate form 0/0
• L'Hôpital's Rule
• d/dx[sin(3x)] = 3cos(3x)
• 3cos(0) = 3

Marks

3

Topic

Maxima and Minima

Difficulty

medium

Template Id

T7

Examiner Tip

Write the classification rule explicitly: 'f″ > 0 → minimum, f″ < 0 → maximum' — examiners want to see that you know the test, not just the result.

Model Answer

Given: f(x) = 2x^3 − 9x^2 + 12x − 4 Required: Critical points and their classification Solution: Step 1 — First Derivative: f′(x) = 6x^2 − 18x + 12 Step 2 — Set f′(x) = 0: 6x^2 − 18x + 12 = 0 x^2 − 3x + 2 = 0 (÷ by 6) (x − 1)(x − 2) = 0 ∴ x = 1 and x = 2 (critical points) Step 3 — Second Derivative Test: f″(x) = 12x − 18 At x = 1: f″(1) = 12(1) − 18 = −6 < 0 → Local MAXIMUM f(1) = 2(1) − 9(1) + 12(1) − 4 = 1 At x = 2: f″(2) = 12(2) − 18 = +6 > 0 → Local MINIMUM f(2) = 2(8) − 9(4) + 12(2) − 4 = 16 − 36 + 24 − 4 = 0 ∴ Local maximum at (1, 1); Local minimum at (2, 0)

Question Type

numerical

Answer Structure

• Step 1: Compute f′(x) = 6x^2 − 18x + 12 correctly [1 mark]
• Step 2: Solve f′(x) = 0 to obtain critical points x = 1 and x = 2 [1 mark]
• Step 3: Apply second-derivative test and correctly classify both points with function values [1 mark]

Scoring Breakdown

Common Mark Deductions

• Finding x values but not computing y = f(x) values for the complete coordinate
• Stating 'minimum' or 'maximum' without the sign of f″ — reasoning must be shown
• Arithmetic error in evaluating f(2): expanding 2(8) = 16, not 2(4)

Key Phrases To Include

• f′(x) = 0
• critical points
• second derivative test
• f″(1) < 0 → local maximum
• f″(2) > 0 → local minimum

Marks

5

Topic

Maxima and Minima — Optimization

Difficulty

hard

Template Id

T8

Examiner Tip

Label the constraint equation and objective function explicitly. Examiners follow a mark scheme; each labelled step corresponds to a mark. A well-organized solution earns all 5 marks even with a minor arithmetic slip.

Model Answer

Given: Total fencing available = 80 m; one side provided by existing wall. Required: Dimensions (length × width) for maximum area. Solution: Step 1 — Define variables: Let x = width (perpendicular to wall), m Let y = length (parallel to wall), m Constraint: 2x + y = 80 Step 2 — Express constraint: y = 80 − 2x Step 3 — Write objective function (area): A(x) = x · y = x(80 − 2x) = 80x − 2x^2 Step 4 — Differentiate and set to zero: A′(x) = 80 − 4x = 0 x = 20 m Step 5 — Second-derivative test: A″(x) = −4 < 0 → Maximum confirmed ✓ Step 6 — Find y: y = 80 − 2(20) = 40 m Step 7 — Maximum area: A_max = 20 × 40 = 800 m² ∴ The dimensions for maximum area are 20 m × 40 m, giving A_max = 800 m².

Question Type

long_answer

Answer Structure

• Step 1: Define variables clearly with units [0.5 mark]
• Step 2: Write constraint equation 2x + y = 80 [0.5 mark]
• Step 3: Form objective function A(x) = 80x − 2x^2 in one variable [1 mark]
• Step 4: Differentiate and solve A′(x) = 0 → x = 20 m [1 mark]
• Step 5: Confirm maximum via second-derivative test A″ = −4 < 0 [1 mark]
• Step 6–7: State both dimensions and maximum area with units [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using y = 80 − x instead of y = 80 − 2x (forgetting both widths use fencing)
• Not applying the second-derivative test — loses the classification mark
• Omitting units (metres, m²) on final answer
• Solving for x but not computing y — incomplete answer
• Expressing A in two variables without substituting the constraint first

Key Phrases To Include

• constraint
• objective function
• A′(x) = 0
• x = 20 m
• y = 40 m
• A″ < 0 → maximum
• A_max = 800 m²

Marks

5

Topic

Maxima and Minima — Engineering Optimization

Difficulty

hard

Template Id

T9

Examiner Tip

The key phrase 'open cylindrical tank (no top)' determines the surface area formula. Read the problem carefully and write 'No top: S = πr^2 + 2πrh' in the solution — this clarifies your formula choice to the examiner.

Model Answer

Given: Volume V = 16π m³; open-top cylinder (no lid). Required: Radius r and height h for minimum surface area. Solution: Step 1 — Objective function (surface area — base + lateral): S = πr^2 + 2πrh Step 2 — Volume constraint: V = πr^2 h = 16π ∴ h = 16/r^2 Step 3 — Substitute h into S: S(r) = πr^2 + 2πr·(16/r^2) S(r) = πr^2 + 32π/r Step 4 — Differentiate and set to zero: S′(r) = 2πr − 32π/r^2 = 0 2πr = 32π/r^2 r^3 = 16 r = 16^(1/3) = 2∛2 ≈ 2.52 m Step 5 — Second-derivative test: S″(r) = 2π + 64π/r^3 Since S″ > 0 for all r > 0 → Minimum confirmed ✓ Step 6 — Find h: h = 16/r^2 = 16/(16^(2/3)) = 16^(1/3) = 2∛2 ≈ 2.52 m Step 7 — Minimum surface area: S_min = π(16^(2/3)) + 32π/(16^(1/3)) S_min = π·16^(2/3) + 2π·16^(2/3) = 3π·16^(2/3) ≈ 59.9 m² ∴ Minimum surface area at r = h = 2∛2 m ≈ 2.52 m (r = h relationship for open cylinder).

Question Type

long_answer

Answer Structure

• Step 1: Write surface area formula for open cylinder S = πr^2 + 2πrh [0.5 mark]
• Step 2: Use volume constraint to express h = 16/r^2 [1 mark]
• Step 3: Substitute into S to get S(r) = πr^2 + 32π/r [1 mark]
• Step 4: Differentiate, set S′(r) = 0, solve for r = 16^(1/3) [1 mark]
• Step 5: Apply second-derivative test to confirm minimum [0.5 mark]
• Step 6–7: Find h and state S_min with units [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using S = 2πr^2 + 2πrh (closed cylinder formula — adds the top lid incorrectly)
• Not eliminating h before differentiating — S must be in one variable only
• Solving r^3 = 16 as r = 16/3 (arithmetic/algebra error)

Key Phrases To Include

• open cylinder
• S = πr^2 + 2πrh
• constraint πr^2h = 16π
• h = 16/r^2
• S′(r) = 0
• r = 16^(1/3)
• minimum confirmed

Marks

5

Topic

Related Rates

Difficulty

hard

Template Id

T10

Examiner Tip

Underline or circle the step 'Differentiate w.r.t. t FIRST, substitute values SECOND' on your answer sheet — this is the cardinal rule of related rates and examiners check it.

Model Answer

Given: Ladder length L = 5 m (constant); dx/dt = +0.4 m/s; x = 3 m at the instant of interest. Required: dy/dt when x = 3 m. Solution: Step 1 — Draw diagram and label: x = horizontal distance (base to wall), y = vertical height (top on wall). Pythagorean relation: x^2 + y^2 = L^2 = 25 Step 2 — Find y when x = 3: y = √(25 − 9) = √16 = 4 m Step 3 — Differentiate the relation with respect to t: d/dt(x^2 + y^2) = d/dt(25) 2x(dx/dt) + 2y(dy/dt) = 0 Step 4 — Substitute known values: 2(3)(0.4) + 2(4)(dy/dt) = 0 2.4 + 8(dy/dt) = 0 dy/dt = −2.4/8 = −0.3 m/s ∴ The top of the ladder is sliding DOWN the wall at 0.3 m/s. (Negative sign confirms downward direction — consistent with physics.)

Question Type

long_answer

Answer Structure

• Step 1: Draw and label diagram; write Pythagorean relation x^2 + y^2 = 25 [1 mark]
• Step 2: Calculate y = 4 m when x = 3 m [0.5 mark]
• Step 3: Differentiate x^2 + y^2 = 25 with respect to t to get 2x(dx/dt) + 2y(dy/dt) = 0 [1.5 marks]
• Step 4: Substitute x = 3, y = 4, dx/dt = 0.4 and solve for dy/dt = −0.3 m/s [1 mark]
• Step 5: Interpret the negative sign — sliding down [1 mark]

Scoring Breakdown

Common Mark Deductions

• Substituting x = 3, y = 4 BEFORE differentiating — must differentiate the equation first
• Differentiating with respect to x instead of t
• Not finding y = 4 m first; substituting y = unknown into the differentiated equation
• Omitting the negative sign or failing to interpret it as 'downward'

Key Phrases To Include

• x^2 + y^2 = 25
• differentiate with respect to t
• implicit differentiation
• 2x(dx/dt) + 2y(dy/dt) = 0
• dy/dt = −0.3 m/s
• sliding down

Marks

5

Topic

Related Rates — Conical Tank

Difficulty

hard

Template Id

T11

Examiner Tip

Draw a cross-section of the cone and mark the similar triangles explicitly — this diagram alone earns the method mark for the r–h relationship.

Model Answer

Given: dV/dt = −2 m³/min (draining); Tank: H = 6 m, R = 3 m; h = 4 m at instant of interest. Required: dh/dt Solution: Step 1 — Similar triangles to relate r and h: r/h = R/H = 3/6 = 1/2 → r = h/2 Step 2 — Volume of cone in terms of h only: V = (1/3)πr^2h = (1/3)π(h/2)^2·h = πh^3/12 Step 3 — Differentiate with respect to t: dV/dt = (π/12)·3h^2·(dh/dt) = πh^2/4 · (dh/dt) Step 4 — Substitute known values (h = 4, dV/dt = −2): −2 = π(4)^2/4 · (dh/dt) −2 = π(16)/4 · (dh/dt) −2 = 4π(dh/dt) dh/dt = −2/(4π) = −1/(2π) ≈ −0.159 m/min ∴ The water level is dropping at 1/(2π) ≈ 0.159 m/min when h = 4 m.

Question Type

long_answer

Answer Structure

• Step 1: Similar-triangles proportion r = h/2 [1 mark]
• Step 2: Volume purely in h: V = πh^3/12 [1 mark]
• Step 3: Differentiate w.r.t. t: dV/dt = πh^2/4 · dh/dt [1.5 marks]
• Step 4: Substitute and solve dh/dt = −1/(2π) m/min [1 mark]
• Step 5: State answer with magnitude and direction (dropping) and units [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using V = (1/3)πR^2H with fixed R instead of variable r — must use similar triangles
• Substituting h = 4 into V before differentiating
• Forgetting the chain rule when differentiating h^3 → 3h^2(dh/dt)

Key Phrases To Include

• similar triangles
• r = h/2
• V = πh^3/12
• dV/dt = πh^2/4 · dh/dt
• dh/dt = −1/(2π) m/min

Marks

3

Topic

Tangent Line

Difficulty

medium

Template Id

T12

Examiner Tip

A 3-step presentation (point → slope → line equation) is the standard structure. Each step is a mark — never skip directly to the final equation.

Model Answer

Given: y = x^3 − 2x + 1; x₀ = 2 Required: Equation of tangent line at x = 2 Solution: Step 1 — Find the point (x₀, y₀): y(2) = (2)^3 − 2(2) + 1 = 8 − 4 + 1 = 5 Point: (2, 5) Step 2 — Find the slope m = y′(x₀): y′ = 3x^2 − 2 m = y′(2) = 3(4) − 2 = 12 − 2 = 10 Step 3 — Write tangent line equation (point-slope form): y − y₀ = m(x − x₀) y − 5 = 10(x − 2) y = 10x − 20 + 5 ∴ Tangent line: y = 10x − 15

Question Type

numerical

Answer Structure

• Step 1: Compute y(2) = 5 to get the point (2, 5) [1 mark]
• Step 2: Differentiate and evaluate y′(2) = 10 for the slope [1 mark]
• Step 3: Apply point-slope form and simplify to y = 10x − 15 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using the normal line formula (negative reciprocal slope) instead of tangent
• Evaluating y at the wrong x value
• Not substituting back to get y explicitly — leaving in point-slope form without simplifying

Key Phrases To Include

• tangent line
• slope = y′(x₀)
• point-slope form
• y − y₀ = m(x − x₀)
• y = 10x − 15

Marks

3

Topic

Radius of Curvature

Difficulty

medium

Template Id

T13

Examiner Tip

Write the full formula first before substituting — the formula alone earns the first mark. A clean substitution on the second line earns the second mark.

Model Answer

Given: y = x^2; Point: (0, 0) Required: Radius of curvature R Formula: R = [1 + (y′)^2]^(3/2) / |y″| Solution: Step 1 — First derivative: y′ = 2x → y′(0) = 0 Step 2 — Second derivative: y″ = 2 → y″(0) = 2 Step 3 — Substitute into curvature formula: R = [1 + (0)^2]^(3/2) / |2| R = [1]^(3/2) / 2 R = 1/2 ∴ R = 0.5 m (or 1/2 units, depending on the unit of x)

Question Type

numerical

Answer Structure

• Step 1: State the radius of curvature formula R = [1 + (y′)^2]^(3/2) / |y″| [1 mark]
• Step 2: Compute y′ = 2x → y′(0) = 0 and y″ = 2 [1 mark]
• Step 3: Substitute and evaluate R = 1/2 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Forgetting the absolute value on y″ — must write |y″|
• Applying the formula without evaluating y′ and y″ at the specific point
• Writing R = [1 + (y′)^2]^(1/2) / |y″| — exponent 3/2 not 1/2

Key Phrases To Include

• R = [1 + (y′)^2]^(3/2) / |y″|
• y′ = 2x
• y″ = 2
• R = 1/2

Marks

2

Topic

Partial Derivatives

Difficulty

medium

Template Id

T14

Examiner Tip

Always write 'treat y as constant' and 'treat x as constant' as notes beside each partial derivative — this communicates your understanding of partial differentiation to the examiner.

Model Answer

Given: z = 3x^2y + 2xy^3 − 5y Required: ∂z/∂x and ∂z/∂y Solution: For ∂z/∂x (treat y as constant): ∂z/∂x = 6xy + 2y^3 For ∂z/∂y (treat x as constant): ∂z/∂y = 3x^2 + 6xy^2 − 5 ∴ ∂z/∂x = 6xy + 2y^3; ∂z/∂y = 3x^2 + 6xy^2 − 5

Question Type

numerical

Answer Structure

• Step 1: ∂z/∂x treating y constant → 6xy + 2y^3 [1 mark]
• Step 2: ∂z/∂y treating x constant → 3x^2 + 6xy^2 − 5 [1 mark]

Scoring Breakdown

Common Mark Deductions

• Differentiating 5y with respect to x and writing −5 instead of 0
• Treating both variables as functions when finding a partial derivative
• Forgetting that 3x^2y differentiates to 3x^2 (not 6xy) with respect to y

Key Phrases To Include

• treat y as constant
• treat x as constant
• ∂z/∂x = 6xy + 2y^3
• ∂z/∂y = 3x^2 + 6xy^2 − 5

Marks

3

Topic

Derivative Rules — Combined Rules

Difficulty

hard

Template Id

T15

Examiner Tip

For compound derivatives, use a three-label setup: u = ..., v = ..., u′ = ..., v′ = ... before writing the Product Rule. This structure is worth at least 1 method mark even if final simplification has a minor error.

Model Answer

Given: y = ln(sin x) · e^(2x) Required: dy/dx Solution: Let u = ln(sin x) and v = e^(2x) Step 1 — Differentiate u using Chain Rule: u′ = 1/sin(x) · cos(x) = cos(x)/sin(x) = cot(x) Step 2 — Differentiate v using Chain Rule: v′ = e^(2x) · 2 = 2e^(2x) Step 3 — Apply Product Rule y′ = u′v + uv′: y′ = cot(x) · e^(2x) + ln(sin x) · 2e^(2x) Step 4 — Factor e^(2x): y′ = e^(2x) [cot(x) + 2·ln(sin x)] ∴ dy/dx = e^(2x) [cot x + 2 ln(sin x)]

Question Type

numerical

Answer Structure

• Step 1: Correct u′ = cot x using chain rule on ln(sin x) [1 mark]
• Step 2: Correct v′ = 2e^(2x) using chain rule on e^(2x) [0.5 mark]
• Step 3: Correct product rule application u′v + uv′ [1 mark]
• Step 4: Simplified factored form [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• d/dx[ln(sin x)] = 1/sin x without applying chain rule — missing cos x factor
• Applying only product rule without chain rule on the inner functions
• Not factoring e^(2x) — leaving answer unfully simplified

Key Phrases To Include

• Product Rule
• Chain Rule
• d/dx[ln(sin x)] = cot x
• d/dx[e^(2x)] = 2e^(2x)
• y′ = e^(2x)[cot x + 2ln(sin x)]