Let h(x) = 5g(x) - 3x^2 + 2 sin(x) - 7x, and suppose g'(0) = 2, find h'(0)

Answer 1

well, what's h'(x)? Use the chain rule:

h'(x) = 5g'(x) - 6x + 2sinx - 7
h'(2) = 5g'(2) - 6(2) + 2sin(2) - 7
= 5*2 - 6*2 + 2sin(2) - 7
= 2sin(2) - 9

Answer 2

Ahem: h'(0) = 5*2-0+0-7 = 3

Answer 3

it asks for h'(0) not h'(2). I got the answer 5 but the key says the answer is 1

Answer 4

isn't it 2cos(x)

Answer 5

good catch. You are correct
h'(0) = 10-0+2-7 = 5

what was I thinking?

the answer is 1 if it was supposed to be -2sin(x) instead of +2sin(x)