Suppose for a differentiable function h, h(0) = 0, h(1) = 1 and h'(0) = h'(1) = 2. If g(x) = h (ex )eh(x), then g'(0) is equal to:
- 4
- 5
- 3
- 8
Answer
(1) 4
Solution
Product rule + chain rule:
g'(x) = h' (ex ) ex eh(x) + h (ex ) eh(x) h'(x)
At x = 0: e0 = 1 and eh(0) = e0 = 1
g'(0) = h'(1) 1 1 + h(1) 1 h'(0)
= 2 + 1 2 = 4
Answer: (1)