Differentiate from first principles
12th grade (Year 13 in England) · 16 questions · Common Core: HSF.IF.B.6 · England: A-level Pure Mathematics
Name
Date
Score/ 16
Question 1
Using first principles, f(x) = −2x² + x gives (f(x + h) − f(x))/h = −4x − 2h + 1. What is f′(x)?
Answer
Question 2
For f(x) = x² − 6x, simplify (f(x + h) − f(x))/h.
- A. 2x − 6
- B. 2xh + h²
- C. 2x + h² − 6
- D. 2x + h − 6
Question 3
Using first principles, f(x) = −5x² + 3x gives (f(x + h) − f(x))/h = −10x − 5h + 3. What is f′(x)?
Answer
Question 4
For f(x) = x² − 6x, simplify (f(x + h) − f(x))/h.
- A. 2x − 6
- B. 2x + h − 6
- C. x² + h − 6
- D. 2x + h² − 6
Question 5
Using first principles, f(x) = 4x² − 5x gives (f(x + h) − f(x))/h = 8x + 4h − 5. What is f′(x)?
Answer
Question 6
For f(x) = x² + 4x, simplify (f(x + h) − f(x))/h.
- A. 2x + h + 4
- B. x² + h + 4
- C. 2x + h² + 4
- D. 2x + 4
Question 7
Using first principles, f(x) = −2x² + 2x gives (f(x + h) − f(x))/h = −4x − 2h + 2. What is f′(x)?
Answer
Question 8
For f(x) = x² + 2x, simplify (f(x + h) − f(x))/h.
- A. 2x + h + 2
- B. x² + h + 2
- C. 2xh + h²
- D. 2x + 2
Question 9
Using first principles, f(x) = −2x² − 3x gives (f(x + h) − f(x))/h = −4x − 2h − 3. What is f′(x)?
Answer
Question 10
For f(x) = x² − 4x, simplify (f(x + h) − f(x))/h.
- A. 2x − 4
- B. 2xh + h²
- C. 2x + h − 4
- D. x² + h − 4
Question 11
Using first principles, f(x) = −2x² − 3x gives (f(x + h) − f(x))/h = −4x − 2h − 3. What is f′(x)?
Answer
Question 12
For f(x) = x² + 6x, simplify (f(x + h) − f(x))/h.
- A. x² + h + 6
- B. 2xh + h²
- C. 2x + h + 6
- D. 2x + h² + 6
Question 13
Using first principles, f(x) = −2x² + x gives (f(x + h) − f(x))/h = −4x − 2h + 1. What is f′(x)?
Answer
Question 14
For f(x) = x² + 3x, simplify (f(x + h) − f(x))/h.
- A. 2x + 3
- B. 2x + h + 3
- C. 2x + h² + 3
- D. x² + h + 3
Question 15
Using first principles, f(x) = 3x² − 2x gives (f(x + h) − f(x))/h = 6x + 3h − 2. What is f′(x)?
Answer
Question 16
For f(x) = x², simplify (f(x + h) − f(x))/h.
- A. 2x + h²
- B. 2x + h
- C. 2xh + h²
- D. x² + h
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