Derivative of cos2x/1-sin2x: (2cos^3x - 2sin^2x) / (1-2sin2x + sin^4x)
The derivative of cos2x/(1-sin2x) can be found using the quotient rule:
d/dx [f(x)/g(x)] = [f'(x)g(x) - f(x)g'(x)] / [g(x)]^2
Here, f(x) = cos2x and g(x) = 1-sin2x.
First, find the derivatives of f and g:
f'(x) = -2sin2x
g'(x) = -2cos2x
Now apply the quotient rule:
d/dx [cos2x/(1-sin2x)] = [-2sin2x(1-sin2x) - cos2x(-2cos2x)] / [(1-sin2x)^2]
Simplify the expression:
= (-2sin2x + 2sin^2x + 2cos^3x) / (1-2sin2x + sin^4x)
= (2cos^3x - 2sin^2x) / (1-2sin2x + sin^4x)
This expression can be simplified further using trigonometric identities, but it's already in a valid form for the derivative.
Question:
What is the derivative of cos2x/1-sin2x?