Final answer:
Using the product rule to differentiate g(x) and h(x), we find g'(3) to be -3sin(3) + 4cos(3) and h'(3) to be -4sin(3) - 3cos(3), given that f(3) = 4 and f'(3) = -3.
Step-by-step explanation:
To find the derivatives of the functions g(x) = f(x)sin(x) and h(x) = cos(x)f(x), we apply the product rule. The product rule states that if you have a product of two functions, say u(x)v(x), the derivative u'v + uv' applies. As we know that f(3) = 4 and f'(3) = -3, we can plug these values into the derivatives of g(x) and h(x).
For g'(x), the derivative is f'(x)sin(x) + f(x)cos(x). So, g'(3) is f'(3)sin(3) + f(3)cos(3) = (-3)sin(3) + 4cos(3).
Similarly, for h'(x), the derivative is -sin(x)f(x) + cos(x)f'(x). Therefore, h'(3) is -sin(3)f(3) + cos(3)f'(3) = -sin(3)(4) + cos(3)(-3).
Thus, g'(3) equals -3sin(3) + 4cos(3), and h'(3) equals -4sin(3) - 3cos(3).