To find the derivative of the given function, we can use the product rule and chain rule. Let's break down the process step by step:
Step 1: Apply the product rule
The product rule states that if we have two functions, u(x) and v(x), then the derivative of their product, w(x) = u(x) * v(x), is given by:
w'(x) = u'(x) * v(x) + u(x) * v'(x)
In our case, u(x) = cos x and v(x) = ln(cos^2x). Let's find the derivatives of both terms.
Step 2: Find the derivative of the first term, u(x) = cos x
The derivative of cos x is given by:
u'(x) = -sin x
Step 3: Find the derivative of the second term, v(x) = ln(cos^2x)
We can rewrite v(x) using the property of logarithms: ln(a^b) = b * ln(a)
v(x) = 2 ln(cos x)
To find the derivative of v(x), we apply the chain rule. The chain rule states that if we have a composition of functions, f(g(x)), then the derivative is given by:
(f(g(x)))' = f'(g(x)) * g'(x)
In our case, f(x) = ln x and g(x) = cos x. Let's find the derivatives of both terms.
The derivative of f(x) = ln x is given by:
f'(x) = 1/x
The derivative of g(x) = cos x is given by:
g'(x) = -sin x
Applying the chain rule, we get:
v'(x) = 2 * (1/cos x) * (-sin x)
= -2sin(x)/cos(x)
= -2tan(x)
Step 4: Use the product rule
Now that we have u'(x) and v'(x), we can apply the product rule to find the derivative of the function:
w'(x) = u'(x) * v(x) + u(x) * v'(x)
= -sin x * ln(cos^2x) + cos x * (-2tan(x))
= -sin(x) ln(cos^2x) -2cos(x)tan(x)
Therefore, the derivative of y = cos x * ln(cos^2x) is:
y'(x) = -sin(x) ln(cos^2x) -2cos(x)tan(x)
Now you can use this derivative to solve the problem given in your question.