find trig derivative of y= (2x-cosx^2-sinx^2-xcosx-xsinx-x^2cosx-x^2sinx-x^2)^4

Question

find trig derivative of y= (2x-cosx^2-sinx^2-xcosx-xsinx-x^2cosx-x^2sinx-x^2)^4

 

i'm unsure about the cancellations at the end when you reduce to get the final answer, please help!

Answer 1

Solving this problem is extremely tedious but doable. First we will assume that cos(x²), sin(x²) are typo, they are respectively cos²(x), sin²(x), as noted by Raffael G. above. Otherwise, it will not be solvable. Hence - cos²(x) - sin²(x) = -1 by Pythagorean theorem.

Note also that in the outline below, cos(x) + sin(x) = (2/sqrt 2)[cos(x)sin(45) + sin(x)cos(45)] = (2/sqrt2)sin(x + 45). Here are the outline only:

Let y = z⁴, then y' = (4z³)(z') by chain rule. Here z must first be simplified into the following for easy differentiation:

z = (2x - cos²(x) - sin²(x) - xcos(x) - xsin(x) - x²cos(x) - x²sin(x) - x²)

z = [- (x² - 2x + 1) - cos(x)[x² + x] - sin(x)[x² + x] ]

z = - [(x -1)² + [x² + x] [cos(x) + sin(x)]]

z = - [(x -1)² + (2/sqrt 2)[x² + x] [sin(x + 45)]]

Now that z is made simpler, finding the derivative z' is easier, and hence finding y' is very doable. Hope this outline helps. Good luck!