Advanced differentiation and Maclaurin series
Differentiate inverse and hyperbolic functions, find second derivatives implicitly or parametrically, and derive initial Maclaurin terms.
Section 2.3 · Differentiation
Exam scope and assessment notes
Mapped to Cambridge International AS & A Level Further Mathematics 9231. These original explanations and self-checks are independent study support, not Cambridge questions, marks or grades. The syllabus assumes prior knowledge of the relevant 9709 Mathematics content; check each component's prerequisites with your teacher. Full-paper assessment and human subject-specialist review remain outstanding.
Before you begin
- Cambridge expects learners to know the whole Cambridge International A Level Mathematics (9709) content. Paper 1 further pure mathematics draws on 9709 Papers 1 and 3. Confirm your exam year and component route with your school.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
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What you will learn
- Differentiate hyperbolic and inverse trigonometric or hyperbolic functions
- Find second derivatives for implicit and parametric relations
- Derive and use the first few Maclaurin terms without a general-term derivation
Know the derivative families and their domains
The hyperbolic derivatives are (sinh x)′=cosh x, (cosh x)′=sinh x, (tanh x)′=sech²x, (sech x)′=−sech x tanh x, (cosech x)′=−cosech x coth x and (coth x)′=−cosech²x.
For inverse trigonometric functions, (sin⁻¹x)′=1/√(1−x²) and (cos⁻¹x)′=−1/√(1−x²). For inverse hyperbolic functions, (sinh⁻¹x)′=1/√(1+x²), (cosh⁻¹x)′=1/√(x²−1) for x>1, and (tanh⁻¹x)′=1/(1−x²) for |x|<1.
These follow by differentiating the defining relation and checking its domain. For y=sinh⁻¹x, sinh y=x gives cosh y·dy/dx=1; cosh²y=1+x² supplies the positive square root.
Implicit, parametric and series methods need careful order
For an implicit curve F(x,y)=0, differentiate once to obtain y′, then differentiate again and substitute the relation only when it simplifies the result.
If x=x(t) and y=y(t), dy/dx=(dy/dt)/(dx/dt). Differentiate this expression with respect to t and divide by dx/dt again to get d²y/dx². Check that dx/dt is non-zero at the point.
A Maclaurin series starts with f(0)+f′(0)x+f″(0)x²/2! and so on. Repeated implicit differentiation can generate the first few terms; a general formula is not required by this syllabus.
Pause and explain
Which habit best checks your work on advanced differentiation and maclaurin series?
Worked example
Find dy/dx for y=sinh⁻¹x, then find d²y/dx².
Show the worked solution
- Differentiating sinh y=x gives cosh y·y′=1, hence y′=1/cosh y=1/√(1+x²).
- Differentiate y′=(1+x²)⁻¹/².
- y″=−x(1+x²)⁻³/².
Answer y′=1/√(1+x²), y″=−x/(1+x²)³ᐟ².
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Differentiate y=tanh⁻¹(2x) with respect to x, stating the domain.
Give me a hint
Use d(tanh⁻¹u)/du=1/(1−u²) and apply the chain rule.
Compare my reasoning
- Let u=2x, so du/dx=2.
- dy/dx=2/(1−(2x)²).
- The real domain requires |2x|<1.
dy/dx=2/(1−4x²), for −1/2<x<1/2.
Look for these in your work
- Inverse derivative used
- Chain factor included
- Domain stated
For x=t² and y=t³, find d²y/dx² when t>0.
Give me a hint
First find dy/dx, then differentiate that ratio with respect to t and divide by dx/dt.
Compare my reasoning
- dx/dt=2t and dy/dt=3t², so dy/dx=3t/2.
- Differentiate with respect to t: d/dt(3t/2)=3/2.
- Divide by dx/dt=2t.
d²y/dx²=3/(4t).
Look for these in your work
- First derivative correct
- Ratio differentiated with respect to t
- Second division by dx/dt included
Derive the first three non-zero terms of the Maclaurin series for tan x from y′=1+y².
Give me a hint
Assume y=ax+bx³+cx⁵+… and compare powers after differentiating.
Compare my reasoning
- Since tan 0=0, write y=ax+bx³+cx⁵+…; y′=a+3bx²+5cx⁴+… .
- Comparing constants in y′=1+y² gives a=1; comparing x² terms gives 3b=a², so b=1/3.
- Comparing x⁴ terms gives 5c=2ab=2/3, so c=2/15.
tan x=x+x³/3+2x⁵/15+… near x=0.
Look for these in your work
- Series begins at zero
- Powers compared consistently
- Three coefficients correct
Common mistakes
- Forgetting the chain rule for an inverse function
- Dividing by dx/dt only once when finding a parametric second derivative
- Including a general Maclaurin term when only the first few terms are required
What can you explain now?
If x=x(t) and y=y(t), what is dy/dx?
Compare with the explanation
(dy/dt)/(dx/dt)
The chain rule divides the rates of change with respect to the shared parameter.
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Make it stick
How does the chain rule appear in inverse, implicit and parametric differentiation?
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