How do you find the derivative of f(x)=sin2x/cos2x

easy way:

sin2x/cos2x = tan2x
so the derivative is just

2sec^2(2x)

hard way, using the quotient rule:

[2cos(2x)cos(2x) - sin(2x)(-2sin(2x))]/cos^2(2x)

2(cos^2(2x)+sin^2(2x))/cos^2(2x)

2/cos^2(2x)

2sec^2(2x)

To find the derivative of the function f(x) = sin(2x) / cos(2x), we can use the quotient rule.

The quotient rule states that if we have a function in the form of f(x) = g(x) / h(x), where g(x) and h(x) are differentiable functions, then the derivative of f(x) is given by:

f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2

In our case, g(x) = sin(2x) and h(x) = cos(2x).

Now, let's find the derivatives of g(x) and h(x):

g'(x) = d/dx(sin(2x)) = 2cos(2x) (using the chain rule)

h'(x) = d/dx(cos(2x)) = -2sin(2x) (using the chain rule)

Now, we can substitute these values into the quotient rule formula:

f'(x) = (2cos(2x) * cos(2x) - sin(2x) * (-2sin(2x))) / (cos(2x))^2

Simplifying this expression, we get:

f'(x) = (2cos^2(2x) + 2sin^2(2x)) / cos^2(2x)

Using the trigonometric identity cos^2(x) + sin^2(x) = 1, we can simplify further:

f'(x) = (2(1)) / cos^2(2x)

Finally, simplifying the expression, we get:

f'(x) = 2 / cos^2(2x)

Therefore, the derivative of f(x) = sin(2x) / cos(2x) is f'(x) = 2 / cos^2(2x).