Derivative of ln(√(cos2x))

how do you find the derivative of ln(√cos(2x))?

1/sqrt(cos(2x)) *( 1/2)(cos(2x)^(-1/2) * -sin(2x) * 2

= -sin2x/cos(2x) = -tan(2x)

Jen

Use the chain rule, 3 times:

d/dx 2x = 2
=> d/dx cos(2x) = 2 (-sin 2x) = -2 sin 2x
=> d/dx √ cos(2x) = -2 sin 2x (1/2) (cos 2x)^-1/2 = -sin 2x / √cos 2x
=> d/dx ln √ cos(2x) = (-sin 2x / √cos 2x) / √cos 2x ) = -sin 2x / cos 2x = -tan 2x

y = ln(√(cos(2x)) = 1/2*ln(cos(2x))

dy/dx = 1/2*1/(cos(2x)) * (-sin(2x)*2) = -tan(2x)

y = (1/2)ln(cos2x)

rule
y = ln(f(x)) ---> y ' = f '(x)/f(x)

oihgiur