what is the slope of the tangent to the graph of f(x)= ln [cos(x)] at x= pi/6 ???
slope?
f'(x)= 1/cosx * -sinx
or -tanx what is -tan(PI/6) ?
thank you!
To find the slope of the tangent to the graph of the function f(x) = ln[cos(x)] at x = π/6, we can use the concept of differentiation.
Here's how you can calculate it step by step:
Step 1: Find the derivative of the function f(x) = ln[cos(x)]. The derivative of ln(u), where u is a function of x, is given by (1/u) * du/dx. In this case, u = cos(x). Therefore, the derivative of ln[cos(x)] is (1/cos(x)) * (-sin(x)) = -sin(x)/cos(x) = -tan(x).
Step 2: Substitute x = π/6 into the derivative obtained in Step 1. We get -tan(π/6).
Step 3: Simplify the value of -tan(π/6). The tangent of π/6 is 1/√3. Hence, the slope of the tangent to the graph of f(x) = ln[cos(x)] at x = π/6 is -1/√3.
Therefore, the slope of the tangent to the graph of f(x) = ln[cos(x)] at x = π/6 is -1/√3.