Let f(x)= 2ln(cosx), find f''(x)
f ' (x) = (2/cosx)(-sinx) = -2tanx
f '' (x) = - 2sec^2 x
To find the second derivative of f(x) = 2ln(cosx), we need to find the first derivative and then find the derivative of that.
Step 1: Find the first derivative (f'(x)) of f(x):
To find the derivative of ln(cosx), we can use the chain rule. Let's denote u = cos(x):
f'(x) = 2 * d/dx[ln(u)]
Using the chain rule, we have:
f'(x) = 2 * (1/u) * du/dx
Substituting u = cos(x), we have:
f'(x) = 2 * (1/cos(x)) * (-sin(x))
Simplifying, we get:
f'(x) = -2tan(x)
Step 2: Find the second derivative (f''(x)) of f(x):
To find the derivative of -2tan(x), we can use the quotient rule. Let's denote u = -2 and v = tan(x):
f''(x) = d/dx[u/v]
Using the quotient rule, we have:
f''(x) = (v * du/dx - u * dv/dx) / (v^2)
Substituting u = -2 and v = tan(x), we have:
f''(x) = (tan(x) * 0 - (-2) * sec^2(x)) / (tan^2(x))
Simplifying, we get:
f''(x) = 2sec^2(x) / tan^2(x)
Since sec^2(x) = 1 + tan^2(x), we can simplify further:
f''(x) = 2(1 + tan^2(x)) / tan^2(x)
Simplifying even more, we get:
f''(x) = 2 + 2/tan^2(x)
Hence, the second derivative of f(x) = 2ln(cosx) is f''(x) = 2 + 2/tan^2(x).
To find the second derivative of f(x) = 2ln(cosx), we will need to apply the chain rule twice.
The chain rule states that if we have a function f(g(x)), then its derivative is given by f'(g(x)) * g'(x).
Let's start by finding the first derivative, f'(x):
f(x) = 2ln(cosx)
To differentiate ln(cosx), we need to apply the chain rule. Let g(x) = cosx.
So, g'(x) = -sinx.
Now, f'(x) = 2 * (1/cosx) * -sinx = -2tanx.
Now that we have f'(x), we can proceed to find the second derivative, f''(x):
Applying the chain rule to f'(x) = -2tanx, we need to differentiate -2tanx.
Let g(x) = -2tanx, so g'(x) = -2sec^2(x).
Using the chain rule, we have:
f''(x) = g'(x) * (-2)sec^2(x) = -4sec^2(x)sec^2(x) = -4sec^4(x).
Therefore, f''(x) = -4sec^4(x).