F=tan^(-1)(cos(2x))
F'=
Did you type this question properly? Should it be tan(something) :/ Did you mean tan^-1(x)?
tan^-1(cos(2x))
To find the derivative of the given function F, which is F = tan^(-1)(cos(2x)), we can use the chain rule.
The chain rule states that if we have a composition of functions, say F(g(x)), then the derivative of F with respect to x is given by F'(x) = F'(g(x)) * g'(x).
In this case, our outer function is tan^(-1) and our inner function is cos(2x). Let's differentiate these functions separately:
1. Derivative of tan^(-1):
The derivative of tan^(-1)(u) with respect to u is given by 1/(1+u^2). So, the derivative of tan^(-1)(cos(2x)) with respect to cos(2x) is 1/(1+(cos(2x))^2).
2. Derivative of cos(2x):
The derivative of cos(u) with respect to u is given by -sin(u). So, the derivative of cos(2x) with respect to x is -2sin(2x).
Now, we can use the chain rule to find the derivative of the entire function F:
F' = (1/(1+(cos(2x))^2)) * (-2sin(2x))
= -2sin(2x)/(1+(cos(2x))^2)
Therefore, the derivative of F with respect to x is -2sin(2x)/(1+(cos(2x))^2).