If y = logₑ(1 + x + x² + x³), then at x = 0, y′ is:
If y = logₑ(1 + x + x² + x³), then at x = 0, y′ is:
- A. 1
- B. 0
- C. −1
- D. 2
Answer: A) 1
Explanation: y = log((1+x)(1+x²)). y′ = 1/(1+x) + 2x/(1+x²). At x=0, y′ = 1 + 0 = 1.
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