Let h(x) = 5g(x) - 3x^2 + 2 sin(x) - 7x, and suppose g'(0) = 2, find h'(0)
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
Ahem: h'(0) = 5*2-0+0-7 = 3
it asks for h'(0) not h'(2). I got the answer 5 but the key says the answer is 1
isn't it 2cos(x)
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)
To find the derivative of h(x), we need to find the derivative of each term separately and then combine them.
Let's find the derivative of each term:
The derivative of 5g(x) with respect to x is 5 times the derivative of g(x). Since we know g'(0) = 2, the derivative of 5g(x) is 5 times 2, which equals 10.
The derivative of -3x^2 with respect to x is -6x.
The derivative of 2sin(x) with respect to x is 2cos(x).
The derivative of -7x with respect to x is -7.
Now let's combine all the derivatives:
h'(x) = 10 - 6x + 2cos(x) - 7.
To find h'(0), we substitute x=0 into the derivative:
h'(0) = 10 - 6(0) + 2cos(0) - 7.
Simplifying further:
h'(0) = 10 + 0 + 2 - 7.
Finally, we get:
h'(0) = 5.