Let the graph of the function f⁡x = cos⁡x⁢sin⁡x. The equation of a line tangent to f⁡x at x=π/3 is exactly..

Question

y=? The answer should be in slope intercept form.... y=mx+b

Answer 1

The derivative of the function f(x)=cos(x)sin(x) is
f'(x)=-sin^2 (x)+ cos^2 (x)
f'(x)=1-2*(sin(x))^2
So
f'(Pi/3)=1-2*(3/4)
f'(Pi/3)= -1/2,
which serves as the 'm' in your linear equation.
You can determine b by using the point
(x,y)=(Pi/3, f(Pi/3))=(Pi/3, 0.25*sqrt(3))
in your linear equation.
sqrt(3)/4=(-1/2)*(Pi/3)+b
b= -3*sqrt(3)/(2*Pi)
So your linear equation is
y=(-1/2)*x - 3*sqrt(3)/(2*Pi)

Answer 2

f(x) = 1/2 sin 2x so f´(x) = cos 2x and f´(pi/3) = =-1/2
so
y-1/4sqrt(3)=-1/2(x-pi/3)