Critical Points are where the derivative is equal to 0.
But we need to find the derivative. In order to do so, I'll split up f(x)f(x) into two parts -PART 1 = (x-3sin(x))^4(x−3sin(x))4 and PART 2 = xcos(x)xcos(x) and I will combine them using the Sum Rule later.
PART 1:
Let's apply the Chain Rule.
d/dx(x-3sin(x))^4=4*(x-3sin(x))^3*(1-3cos(x))ddx(x−3sin(x))4=4⋅(x−3sin(x))3⋅(1−3cos(x))
=4(x-3sin(x))^3(1-3cos(x))=4(x−3sin(x))3(1−3cos(x))
PART 2:
Let's apply the Product Rule.
d/dx xcos(x)=1*cos(x)+x*-sin(x)ddxxcos(x)=1⋅cos(x)+x⋅−sin(x)
=cos(x)-xsin(x)=cos(x)−xsin(x)
therefore f'(x) is:
=4(x-3sin(x))^3(1-3cos(x))-(cos(x)-xsin(x))
=xsin(x)-cos(x)+4(x-3sin(x))^3(1-3cos(x)) (Rearranging and expanding -(cos(x)-xsin(x)))
Now all you have to do is set that equal to 0:
xsin(x)-cos(x)+4(x-3sin(x))^3(1-3cos(x))=0
...
