Calculus Made Easy: Derivative of ln(cos(x)) Using the Chain Rule

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Struggling with derivatives of trigonometric functions? This video simplifies the process of finding the derivative of ln(cos(x)) using the chain rule. We'll break down each step, making it easy for you to understand and apply the concept to other problems. Perfect for calculus students or anyone needing a quick refresher!

d/dx f(g(x)) = f'(g(x)) * g'(x)

where f and g are functions of x. In this case, we have:

f(x) = ln(x) and g(x) = cos(x)

So, we need to find f'(g(x)) and g'(x) in order to evaluate the derivative.

Step 1: Find f'(x)

Using the derivative of the natural logarithm, we have:

f'(x) = 1/x

Step 2: Find g'(x)

Using the derivative of the cosine function, we have:

g'(x) = -sin(x)

Step 3: Find f'(g(x))

Substituting g(x) into the formula for f'(x), we get:

f'(g(x)) = 1/cos(x)

Step 4: Apply the chain rule formula

Using the formula d/dx f(g(x)) = f'(g(x)) * g'(x), we can now calculate the derivative of ln(cos(x)):

d/dx ln(cos(x)) = d/dx f(g(x)) = f'(g(x)) * g'(x)
= (1/cos(x)) * (-sin(x))
= -tan(x)/cos(x)

At x = pi/2, cos(x) = 0, which means that the function is not defined at that point.

To find the equation of the tangent line at x = pi/3, we can first evaluate the derivative at that point:

d/dx ln(cos(x)) = -tan(x)/cos(x)
d/dx ln(cos(pi/3)) = -tan(pi/3)/cos(pi/3)
= -√3/3

So, the slope of the tangent line at x = pi/3 is -√3/3.

Next, we can use the point-slope form of a line to find the equation of the tangent line:

y - f(pi/3) = m(x - pi/3)

where m is the slope and (pi/3, f(pi/3)) is the point on the curve.

Substituting the values we have, we get:

y - ln(cos(pi/3)) = (-√3/3)(x - pi/3)

Simplifying, we get:

y = -√3/3 x + (ln(2) - ln(√3)/2)

Therefore, the equation of the tangent line at x = pi/3 is y = -√3/3 x + (ln(2) - ln(√3)/2).

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