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).