Solve the derivative of expression: tan^2(ln(square root(4e^x+2x^2))) + cot^2(ln(square root(4e^x+2x^2)))

We notice that each term of the sum is a composed
function, so, we'll calculate the derivative of each
term.

We'll note the first term as f(x) and the 2nd term as
g(x).

f(x) = tan^2(ln(square
root(4e^x+2x^2)))

To determine the derivative of f(x),
we'll apply the chain rule:

f'(x) =
2tan{ln[sqrt(4e^x+2x^2)]}*(1/[cos(lnsqrt(4e^x+2x^2)]^2*[1/2sqrt(4e^x+2x^2)]*(4e^x +
4x)

We'll calculate the derivative of
g(x):

g'(x) =
-2cot{ln[sqrt(4e^x+2x^2)]}*(1/[sin(lnsqrt(4e^x+2x^2)]^2*[1/2sqrt(4e^x+2x^2)]*(4e^x +
4x)

Now, we'll add f'(x) +
g'(x):

tan{ln[sqrt(4e^x+2x^2)]}*(1/[cos(lnsqrt(4e^x+2x^2)]^2*[1/sqrt(4e^x+2x^2)]*(4e^x
+ 4x) - cot{ln[sqrt(4e^x+2x^2)]}*(1/[sin(lnsqrt(4e^x+2x^2)]^2*[1/sqrt(4e^x+2x^2)]*(4e^x
+ 4x)

We'll factorize by (4e^x +
4x)/sqrt(4e^x+2x^2)

 f'(x) + g'(x) = [(4e^x +
4x)/sqrt(4e^x+2x^2)]*{tan ln[sqrt(4e^x+2x^2)]/[cos(lnsqrt(4e^x+2x^2)]^2] - cot
ln[sqrt(4e^x+2x^2)]/[sin(lnsqrt(4e^x+2x^2)]^2}