Why is derivative of 4secx+ cotx = 4secxtanx - (cscx)^2 explain

The derivative of secx is secxtanx.
The derivative of cotx is -(cscx)^2

d/dx(4 sec(x) tan(x)-csc^2(x)-4 sec(x)-cot(x)) = (2 cot(x)+1) csc^2(x)-2 sec^3(x) (sin(2 x)+cos(2 x)-3)

Possible derivation:
d/dx(-cot(x)-csc^2(x)-4 sec(x)+4 tan(x) sec(x))
| Differentiate the sum term by term and factor out constants:
= | -d/dx(cot(x))-(d/dx(csc^2(x)))-4 (d/dx(sec(x)))+4 (d/dx(tan(x) sec(x)))
| The derivative of cot(x) is -csc^2(x):
= | -d/dx(csc^2(x))-4 (d/dx(sec(x)))+4 (d/dx(tan(x) sec(x)))--csc^2(x)
| Use the chain rule, d/dx(csc^2(x)) = ( du^2)/( du) ( du)/( dx), where u = csc(x) and ( du^2)/( du) = 2 u:
= | -2 csc(x) (d/dx(csc(x)))-4 (d/dx(sec(x)))+4 (d/dx(tan(x) sec(x)))+csc^2(x)
| The derivative of sec(x) is tan(x) sec(x):
= | -2 csc(x) (d/dx(csc(x)))+4 (d/dx(tan(x) sec(x)))+csc^2(x)-4 (tan(x) sec(x))
| The derivative of csc(x) is -cot(x) csc(x):
= | 4 (d/dx(tan(x) sec(x)))+csc^2(x)-2 csc(x) (-cot(x) csc(x))-4 tan(x) sec(x)
| Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = sec(x) and v = tan(x):
= | 4 (sec(x) (d/dx(tan(x)))+tan(x) (d/dx(sec(x))))+csc^2(x)+2 cot(x) csc^2(x)-4 tan(x) sec(x)
| The derivative of tan(x) is sec^2(x):
= | 4 (tan(x) (d/dx(sec(x)))+sec^2(x) sec(x))+csc^2(x)+2 cot(x) csc^2(x)-4 tan(x) sec(x)
| The derivative of sec(x) is tan(x) sec(x):
= | csc^2(x)+2 cot(x) csc^2(x)+4 (sec^3(x)+tan(x) (tan(x) sec(x)))-4 tan(x) sec(x)