01 Derivative rules and domains Choose power, product, quotient or chain rules from the expression structure, retaining domains. ↗ Question directory1. Foundation · Find f′(2) for f=3x⁴−2x+5. 2. Standard · Find the derivative of (x²+1)(x−2). 3. Transfer · For F(x)=g(x²), F′(2)=24. Find g′(4) and explain why g′(4)=24 is 4. Foundation · Find the derivative of the constant function f(x)=7. 5. Foundation · Find f′(−1) for f=x⁵. 6. Foundation · Find f′(0) for f=(2x+1)³. 7. Foundation · Find f′(2) for f=(x+1)/(x−1). 8. Standard · Differentiate ln(3x−2), stating its real domain. 9. Standard · Find the derivative of sin(2x), where x is in radians. 10. Standard · Find the derivative of e^(−x). 11. Standard · Find f′(1) for f=x²(x−3). 12. Transfer · A student differentiates x·x as 1·1. Find the missing terms and e 13. Transfer · For f(x)=(ax+1)³, find all real a such that f′(0)=12. ← →
TOPIC 01
Derivative rules and domains Choose power, product, quotient or chain rules from the expression structure, retaining domains.
What you will be able to explain Choose power, product, quotient or chain rules from the expression structure, retaining domains. Recover an unknown inner rate and audit missing derivative factors. Differentiate the structure Multiplying or nesting quantities requires tracing where their changes come from.
A product changes through either factor; a composition passes the rate through its inner input.
( u v ) ′ = u ′ v + u v ′ ; ( f ∘ g ) ′ = f ′ ( g ) g ′ (uv)'=u'v+uv';\ (f\circ g)'=f'(g)g' ( uv ) ′ = u ′ v + u v ′ ; ( f ∘ g ) ′ = f ′ ( g ) g ′ A product changes through either factor; a composition passes the rate through its inner input.
Δ ( u v ) = u Δ v + v Δ u + Δ u Δ v \Delta(uv)=u\Delta v+v\Delta u+\Delta u\Delta v Δ ( uv ) = u Δ v + v Δ u + Δ u Δ v A product changes through either factor; a composition passes the rate through its inner input.
( u / v ) ′ = ( u ′ v − u v ′ ) / v 2 (u/v)'=(u'v-uv')/v^2 ( u / v ) ′ = ( u ′ v − u v ′ ) / v 2 When it applies Both component derivatives must exist; quotient denominators stay nonzero. Real logarithms require positive arguments; trigonometric derivatives here use radians. Counterexample: For x·x, multiplying derivatives gives 1, but the true derivative is 2x.
Expression tree, expanded polynomial, inner-rate table and derivative curve. Pause at these common mistakes Omitting the inner derivative. Multiplying or dividing derivatives as if the original arithmetic rules still apply. PREDICT → EXPLORE → EXPLAIN → TRANSFER
Make a prediction, then explore the relationship. Lesson question: For (kx+1)³, predict the derivative at zero when k changes sign.
Model exploration: predict which displayed result changes with the parameters, check the values, and compare the observation with the lesson question.
Numerical data for the diagram; printable Input Function Derivative Missing factor result -1 -1 6 3 0 1 6 3 1 27 54 27
Explain: Compare the correct 3k(kx+1)² with the expression missing k.
Transfer: At k=0, explain why the function is constant and its derivative vanishes.
My prediction, exploration and explanation Try it. Leave your reasoning visible. Use one hint at a time. A correction explains what changed, not just the final answer.
Examples 1–3: study a full demonstration; finish a solution from its first step; then choose a method independently before opening the solution. The remaining exercises may continue after class.
Find f′(2) for f=3x⁴−2x+5. f ( x ) = 3 x 4 − 2 x + 5 f(x)=3x^4-2x+5 f ( x ) = 3 x 4 − 2 x + 5 Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes. Skills and prerequisite lessons Derivative operations Derivative rules and domains Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
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BUILD THE REASONING
Hint 1+ Differentiate term by term; a constant differentiates to zero.
Hint 2+ Start from this intermediate relation.
f ′ = 12 x 3 − 2 f'=12x^3-2 f ′ = 12 x 3 − 2 Worked solution+ Differentiate term by term; a constant differentiates to zero.
Establish and simplify this mathematical relation.
f ′ = 12 x 3 − 2 f'=12x^3-2 f ′ = 12 x 3 − 2 Establish and simplify this mathematical relation.
f ′ ( 2 ) = 96 − 2 = 94 f'(2)=96-2=94 f ′ ( 2 ) = 96 − 2 = 94 Keep the exponent decrease as well as its multiplier.
The requested value is 94.
Checks and common pitfalls: Keep the exponent decrease as well as its multiplier.
Think first. Reveal a hint when the class is ready.
Find the derivative of (x²+1)(x−2). f = ( x 2 + 1 ) ( x − 2 ) f=(x^2+1)(x-2) f = ( x 2 + 1 ) ( x − 2 ) First step provided — complete the reasoning: Each factor contributes a derivative term.
Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes. Skills and prerequisite lessons Derivative operations Derivative rules and domains Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
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BUILD THE REASONING
Hint 1+ Each factor contributes a derivative term.
Hint 2+ Start from this intermediate relation.
f ′ = 2 x ( x − 2 ) + ( x 2 + 1 ) f'=2x(x-2)+(x^2+1) f ′ = 2 x ( x − 2 ) + ( x 2 + 1 ) Worked solution+ Each factor contributes a derivative term.
Establish and simplify this mathematical relation.
f ′ = 2 x ( x − 2 ) + ( x 2 + 1 ) f'=2x(x-2)+(x^2+1) f ′ = 2 x ( x − 2 ) + ( x 2 + 1 ) Establish and simplify this mathematical relation.
f ′ = 3 x 2 − 4 x + 1 f'=3x^2-4x+1 f ′ = 3 x 2 − 4 x + 1 The product of the derivatives would omit both cross contributions.
f′=3x²−4x+1.
Checks and common pitfalls: The product of the derivatives would omit both cross contributions.
Reasoning checklist · self / teacher assessment Each factor contributes a derivative term. Give the intermediate relations and explain the conclusion with its restrictions. The product of the derivatives would omit both cross contributions. Think first. Reveal a hint when the class is ready.
For F(x)=g(x²), F′(2)=24. Find g′(4) and explain why g′(4)=24 is wrong. F ′ ( 2 ) = 24 ; F ( x ) = g ( x 2 ) F'(2)=24;\ F(x)=g(x^2) F ′ ( 2 ) = 24 ; F ( x ) = g ( x 2 ) Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes. Skills and prerequisite lessons Derivative operations Derivative rules and domains Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
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Hint 1+ Undo the inner derivative factor.
Hint 2+ Start from this intermediate relation.
F ′ ( x ) = 2 x g ′ ( x 2 ) F'(x)=2xg'(x^2) F ′ ( x ) = 2 x g ′ ( x 2 ) Worked solution+ Undo the inner derivative factor.
Establish and simplify this mathematical relation.
F ′ ( x ) = 2 x g ′ ( x 2 ) F'(x)=2xg'(x^2) F ′ ( x ) = 2 x g ′ ( x 2 ) Establish and simplify this mathematical relation.
24 = 4 g ′ ( 4 ) ⟹ g ′ ( 4 ) = 6 24=4g'(4)\implies g'(4)=6 24 = 4 g ′ ( 4 ) ⟹ g ′ ( 4 ) = 6 F′ and g′ are derivatives with respect to different inputs.
The requested value is 6.
Checks and common pitfalls: F′ and g′ are derivatives with respect to different inputs.
Think first. Reveal a hint when the class is ready.
Find the derivative of the constant function f(x)=7. Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes. Skills and prerequisite lessons Derivative operations Derivative rules and domains Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
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Hint 1+ No output change occurs for any nonzero input increment.
Hint 2+ Start from this intermediate relation.
[ f ( x + h ) − f ( x ) ] / h = ( 7 − 7 ) / h [f(x+h)-f(x)]/h=(7-7)/h [ f ( x + h ) − f ( x )] / h = ( 7 − 7 ) / h Worked solution+ No output change occurs for any nonzero input increment.
Establish and simplify this mathematical relation.
[ f ( x + h ) − f ( x ) ] / h = ( 7 − 7 ) / h [f(x+h)-f(x)]/h=(7-7)/h [ f ( x + h ) − f ( x )] / h = ( 7 − 7 ) / h Establish and simplify this mathematical relation.
A constant derivative is zero, not the same constant.
The requested value is 0.
Checks and common pitfalls: A constant derivative is zero, not the same constant.
Think first. Reveal a hint when the class is ready.
Find f′(−1) for f=x⁵. Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes. Skills and prerequisite lessons Derivative operations Derivative rules and domains Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
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Hint 1+ Apply the power rule and evaluate the even power.
Hint 2+ Start from this intermediate relation.
Worked solution+ Apply the power rule and evaluate the even power.
Establish and simplify this mathematical relation.
Establish and simplify this mathematical relation.
A negative input does not force a negative derivative.
The requested value is 5.
Checks and common pitfalls: A negative input does not force a negative derivative.
Think first. Reveal a hint when the class is ready.
Find f′(0) for f=(2x+1)³. f = ( 2 x + 1 ) 3 f=(2x+1)^3 f = ( 2 x + 1 ) 3 Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes. Skills and prerequisite lessons Derivative operations Derivative rules and domains Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
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Hint 1+ Differentiate the outside and multiply by the inside derivative.
Hint 2+ Start from this intermediate relation.
f ′ = 3 ( 2 x + 1 ) 2 ( 2 ) f'=3(2x+1)^2(2) f ′ = 3 ( 2 x + 1 ) 2 ( 2 ) Worked solution+ Differentiate the outside and multiply by the inside derivative.
Establish and simplify this mathematical relation.
f ′ = 3 ( 2 x + 1 ) 2 ( 2 ) f'=3(2x+1)^2(2) f ′ = 3 ( 2 x + 1 ) 2 ( 2 ) Establish and simplify this mathematical relation.
Omitting the inner factor gives half the correct rate.
The requested value is 6.
Checks and common pitfalls: Omitting the inner factor gives half the correct rate.
Think first. Reveal a hint when the class is ready.
Find f′(2) for f=(x+1)/(x−1). f = ( x + 1 ) / ( x − 1 ) ; x ≠ 1 f=(x+1)/(x-1);\ x\ne1 f = ( x + 1 ) / ( x − 1 ) ; x = 1 The rational denominator must be nonzero; excluded inputs are not points on its curve. Skills and prerequisite lessons Derivative operations Derivative rules and domains Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
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Hint 1+ Subtract the numerator terms in the quotient-rule order.
Hint 2+ Start from this intermediate relation.
f ′ = [ ( x − 1 ) − ( x + 1 ) ] / ( x − 1 ) 2 f'=[(x-1)-(x+1)]/(x-1)^2 f ′ = [( x − 1 ) − ( x + 1 )] / ( x − 1 ) 2 Worked solution+ Subtract the numerator terms in the quotient-rule order.
Establish and simplify this mathematical relation.
f ′ = [ ( x − 1 ) − ( x + 1 ) ] / ( x − 1 ) 2 f'=[(x-1)-(x+1)]/(x-1)^2 f ′ = [( x − 1 ) − ( x + 1 )] / ( x − 1 ) 2 Establish and simplify this mathematical relation.
Dividing the derivatives would give 1 and is not a quotient rule.
The requested value is -2.
Checks and common pitfalls: Dividing the derivatives would give 1 and is not a quotient rule.
Think first. Reveal a hint when the class is ready.
Differentiate ln(3x−2), stating its real domain. f = ln ( 3 x − 2 ) f=\ln(3x-2) f = ln ( 3 x − 2 ) Use real x>2/3 so 3x−2>0. Skills and prerequisite lessons Derivative operations Derivative rules and domains Scope: School extension · This task develops the school textbook skill beyond the explicitly listed JAE syllabus topics.
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Hint 1+ Retain the positive argument condition alongside the chain rule.
Hint 2+ Start from this intermediate relation.
f ′ = 3 / ( 3 x − 2 ) f'=3/(3x-2) f ′ = 3/ ( 3 x − 2 ) Worked solution+ Retain the positive argument condition alongside the chain rule.
Establish and simplify this mathematical relation.
f ′ = 3 / ( 3 x − 2 ) f'=3/(3x-2) f ′ = 3/ ( 3 x − 2 ) Establish and simplify this mathematical relation.
3 x − 2 > 0 ⟹ x > 2 / 3 3x-2>0\implies x>2/3 3 x − 2 > 0 ⟹ x > 2/3 A nonzero denominator alone is weaker than the logarithm domain.
f′=3/(3x−2), x>2/3.
Checks and common pitfalls: A nonzero denominator alone is weaker than the logarithm domain.
Reasoning checklist · self / teacher assessment Retain the positive argument condition alongside the chain rule. Give the intermediate relations and explain the conclusion with its restrictions. A nonzero denominator alone is weaker than the logarithm domain. Think first. Reveal a hint when the class is ready.
Find the derivative of sin(2x), where x is in radians. f = sin ( 2 x ) f=\sin(2x) f = sin ( 2 x ) The trigonometric variable is measured in radians. Skills and prerequisite lessons Derivative operations Derivative rules and domains Scope: School extension · This task develops the school textbook skill beyond the explicitly listed JAE syllabus topics.
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Hint 1+ Use the radians derivative and the inner factor 2.
Hint 2+ Start from this intermediate relation.
f ′ = cos ( 2 x ) ( 2 ) f'=\cos(2x)(2) f ′ = cos ( 2 x ) ( 2 ) Worked solution+ Use the radians derivative and the inner factor 2.
Establish and simplify this mathematical relation.
f ′ = cos ( 2 x ) ( 2 ) f'=\cos(2x)(2) f ′ = cos ( 2 x ) ( 2 ) Establish and simplify this mathematical relation.
f ′ = 2 cos ( 2 x ) f'=2\cos(2x) f ′ = 2 cos ( 2 x ) Degree-based variables introduce an additional π/180 factor.
f′=2cos(2x).
Checks and common pitfalls: Degree-based variables introduce an additional π/180 factor.
Reasoning checklist · self / teacher assessment Use the radians derivative and the inner factor 2. Give the intermediate relations and explain the conclusion with its restrictions. Degree-based variables introduce an additional π/180 factor. Think first. Reveal a hint when the class is ready.
Find the derivative of e^(−x). Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes. Skills and prerequisite lessons Derivative operations Derivative rules and domains Scope: School extension · This task develops the school textbook skill beyond the explicitly listed JAE syllabus topics.
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Hint 1+ The inside function −x has derivative −1.
Hint 2+ Start from this intermediate relation.
f ′ = e − x ( − 1 ) f'=e^{-x}(-1) f ′ = e − x ( − 1 ) Worked solution+ The inside function −x has derivative −1.
Establish and simplify this mathematical relation.
f ′ = e − x ( − 1 ) f'=e^{-x}(-1) f ′ = e − x ( − 1 ) Establish and simplify this mathematical relation.
The exponential remains positive although its rate here is negative.
f′=−e^(−x).
Checks and common pitfalls: The exponential remains positive although its rate here is negative.
Reasoning checklist · self / teacher assessment The inside function −x has derivative −1. Give the intermediate relations and explain the conclusion with its restrictions. The exponential remains positive although its rate here is negative. Think first. Reveal a hint when the class is ready.
Find f′(1) for f=x²(x−3). f = x 2 ( x − 3 ) f=x^2(x-3) f = x 2 ( x − 3 ) Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes. Skills and prerequisite lessons Derivative operations Derivative rules and domains Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
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Hint 1+ Differentiate both factors or expand first.
Hint 2+ Start from this intermediate relation.
f ′ = 2 x ( x − 3 ) + x 2 f'=2x(x-3)+x^2 f ′ = 2 x ( x − 3 ) + x 2 Worked solution+ Differentiate both factors or expand first.
Establish and simplify this mathematical relation.
f ′ = 2 x ( x − 3 ) + x 2 f'=2x(x-3)+x^2 f ′ = 2 x ( x − 3 ) + x 2 Establish and simplify this mathematical relation.
f ′ ( 1 ) = − 4 + 1 = − 3 f'(1)=-4+1=-3 f ′ ( 1 ) = − 4 + 1 = − 3 Substitute only after the full derivative has been formed.
The requested value is -3.
Checks and common pitfalls: Substitute only after the full derivative has been formed.
Think first. Reveal a hint when the class is ready.
A student differentiates x·x as 1·1. Find the missing terms and explain the error. Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes. Skills and prerequisite lessons Derivative operations Derivative rules and domains Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
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BUILD THE REASONING
Hint 1+ Track a change in either factor while the other remains.
Hint 2+ Start from this intermediate relation.
Δ ( x 2 ) = 2 x Δ x + ( Δ x ) 2 \Delta(x^2)=2x\Delta x+(\Delta x)^2 Δ ( x 2 ) = 2 x Δ x + ( Δ x ) 2 Worked solution+ Track a change in either factor while the other remains.
Establish and simplify this mathematical relation.
Δ ( x 2 ) = 2 x Δ x + ( Δ x ) 2 \Delta(x^2)=2x\Delta x+(\Delta x)^2 Δ ( x 2 ) = 2 x Δ x + ( Δ x ) 2 Establish and simplify this mathematical relation.
f ′ = lim h → 0 ( 2 x + h ) = 2 x f'=\lim_{h\to0}(2x+h)=2x f ′ = h → 0 lim ( 2 x + h ) = 2 x For example x=0 gives derivative 0 rather than 1.
f′=1·x+x·1=2x; derivative products are not the product rule.
Checks and common pitfalls: For example x=0 gives derivative 0 rather than 1.
Reasoning checklist · self / teacher assessment Track a change in either factor while the other remains. Give the intermediate relations and explain the conclusion with its restrictions. For example x=0 gives derivative 0 rather than 1. Think first. Reveal a hint when the class is ready.
For f(x)=(ax+1)³, find all real a such that f′(0)=12. f = ( a x + 1 ) 3 f=(ax+1)^3 f = ( a x + 1 ) 3 Use the stated real domain and distinguish stationary candidates, attained extrema and actual concavity changes. Skills and prerequisite lessons Derivative operations Derivative rules and domains Scope: JM02 · This task uses elementary probability rules or polynomial calculus named in the JAE syllabus.
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BUILD THE REASONING
Hint 1+ Express the requested derivative in terms of the unknown parameter.
Hint 2+ Start from this intermediate relation.
f ′ ( x ) = 3 a ( a x + 1 ) 2 f'(x)=3a(ax+1)^2 f ′ ( x ) = 3 a ( a x + 1 ) 2 Worked solution+ Express the requested derivative in terms of the unknown parameter.
Establish and simplify this mathematical relation.
f ′ ( x ) = 3 a ( a x + 1 ) 2 f'(x)=3a(ax+1)^2 f ′ ( x ) = 3 a ( a x + 1 ) 2 Establish and simplify this mathematical relation.
f ′ ( 0 ) = 3 a = 12 ⟹ a = 4 f'(0)=3a=12\implies a=4 f ′ ( 0 ) = 3 a = 12 ⟹ a = 4 The inner derivative depends on a, not on x alone.
The requested value is 4.
Checks and common pitfalls: The inner derivative depends on a, not on x alone.
Think first. Reveal a hint when the class is ready.
