Let y = ln(cos x ). Which of the following is dy/dx?*1/ln(cos x )cotx-tanx-secx
Question
Let y = ln(cos x ). Which of the following is dy/dx?*1/ln(cos x )cotx-tanx-secx
Solution
To find dy/dx, we need to differentiate y = ln(cos x) with respect to x.
Step 1: Apply the chain rule. The derivative of ln(u) with respect to u is 1/u. So, we have:
dy/dx = (1/cos x) * d(cos x)/dx
Step 2: Differentiate cos x with respect to x. The derivative of cos x is -sin x.
dy/dx = (1/cos x) * (-sin x)
Step 3: Simplify the expression.
dy/dx = -sin x / cos x
Step 4: Use trigonometric identities to simplify further. Recall that sin x / cos x is equal to tan x.
dy/dx = -tan x
Therefore, the correct answer is -tan x.
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