Question

Let f and g be differentiable functions such that f'(0) = 3 and g'(0) = 7. If h(x) = 3f(x) - 2g(x) - 5 cos x - 3, what is the value of h'(0)?

Answer 1

The value of h'(0) is -5.

Step-by-step explanation:

To find the value of h'(0), we can differentiate the function h(x) = 3f(x) - 2g(x) - 5 cos(x) - 3 with respect to x. Let's break down the differentiation process step by step:

1. For 3f(x), the derivative will be 3f'(x). Since we're looking for the value at x = 0, this term will be 3f'(0).
2. For -2g(x), the derivative will be -2g'(x). At x = 0, this term becomes -2g'(0).
3. The derivative of -5 cos(x) is -5 (-sin(x)) = 5 sin(x). Again, at x = 0, this term becomes 5 sin(0) = 0.
4. The derivative of a constant term (-3) is 0.

Summing up the derivative of each term, we get: h'(x) = 3f'(0) - 2g'(0). Substituting the given values, we have h'(0) = 3(3) - 2(7) = 9 - 14 = -5.