Final answer:
The differentiation of the function f(x) = (-cos(x))sin(x) using the product rule gives us f'(x) = 1 - 2cos2(x).
Step-by-step explanation:
The question asks us to differentiate the function f(x) = (-cos(x))sin(x).
To do this, we can apply the product rule of differentiation, which states if you have a function that is the product of two functions, its derivative is given by the first function times the derivative of the second function plus the second function times the derivative of the first function. So:
f'(x) = d/dx[(-cos(x))sin(x)]
Using the product rule:
f'(x) = (-cos(x)) * d/dx[sin(x)] + sin(x) * d/dx[-cos(x)]
Simplifying this, knowing the derivative of sin(x) is cos(x), and the derivative of cos(x) is -sin(x), we get:
f'(x) = (-cos(x))(cos(x)) + sin(x)(sin(x))
f'(x) = -cos2(x) + sin2(x)
Using the Pythagorean identity sin2(x) + cos2(x) = 1, we can further simplify the result to:
f'(x) = -cos2(x) + (1 - cos2(x))
f'(x) = -cos2(x) + 1 - cos2(x)
Finally, we combine like terms:
f'(x) = 1 - 2cos2(x)