The graph of y=cos x * ln cos^2x has seven horizontal tangent lines on the interval [0,2pi]. Find the x-coordinate of all points at which these tangent lines occur.
What the derivative of the function?
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.