What is the derivative of ln(cos^2(x))?

Question

my answer was -tan(x)

Answer 1

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

Answer 2

Let f(x) = ln(cos^2(x)). Then, according to the properties of the logarithmic function, f(x) = 2 ln(cos(x)). Also based on the properties of the logarithmic function, it follows that

f'(x) = 2(d/dx cos(x))/(cos(x)). Since d/dx cos(x) = -sin(x), we have f'(x) = 2 (-sin(x))/(cos(x) = - 2 tan(x)

Answer 3

Hi,

Not quite. With natural logs, you say, 1 over the inner function:

1/cos^2(x)

Then multiply by the derivative of that inner function. But you see the derivative of the inner function would be:

2 cos(x) -sin(x).

So, multiply that expression by the derivative of the outside that I showed you a few lines above:

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

OR you can write it as:

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

I hope that helps you out! Please e-mail or instant message me if you have any other questions!

Sincerely,

Andrew

Answer 4

Let y = ln(cos ² x)
Let u = cos ² x
u = (cos x)²
du/dx = 2(cos x).(- sinx)
du/dx = - 2 cosx.sinx
y = ln u
dy/du = 1 / u
dy/du = 1 / cos ² x
dy/dx = (dy/du) X (du/dx)
dy/dx = (1/cos²x) X (-2.cos x.sin x)
dy/dx = - 2.sin x/cos x
dy/dx = - 2 tan x

Answer 5

-2tan(x)