4PM1/02R Jan 2023, Question 3: The Product and Quotient Rules
Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 2R · Calculator · 8 marks
اقرأ هذا الحل مشروحاً بالعربي ←
The question, in short
For y = e^(3x) sin 2x, differentiate twice and show that 13y + d²y/dx² equals 6 dy/dx.
Summarised in our own words. Use the official paper for the exact wording and any diagrams.
Solution · 8 marks
- 1dy/dx = 3e^(3x) sin 2x + 2e^(3x) cos 2x
- 2d²y/dx² = 3(3e^(3x) sin 2x + 2e^(3x) cos 2x) + 6e^(3x) cos 2x − 4e^(3x) sin 2x
- 3So d²y/dx² = 3 dy/dx + 6e^(3x) cos 2x − 4y
- 4From dy/dx: 2e^(3x) cos 2x = dy/dx − 3y, so 6e^(3x) cos 2x = 3 dy/dx − 9y
- 5d²y/dx² = 3 dy/dx + 3 dy/dx − 9y − 4y
- 613y + d²y/dx² = 6 dy/dx
Answer13y + d²y/dx² = 6 dy/dx shown
Stuck on a step? Read the free lesson: The Product and Quotient Rules
Practise more the product and quotient rules questions
Real 4PM1 past-paper questions, marked on your working like the exam, with a worked solution for every part.
Practise free on LemmaraMore past-paper questions on the product and quotient rules
Also searched as: 4PM1 2R January 2023 · 4PM1/02R · 4pm1 jan23 2r · question 3.
Lemmara is an independent study tool, not affiliated with or endorsed by Pearson Edexcel. The worked solutions are Lemmara’s own.