Problem 10 - Calculus Early Transcendentals

Question

Evaluate the following derivatives. $$\frac{d}{d x}\left(\ln \left(\cos ^{2} x\right)\right)$$

Step-by-Step Solution

Answer

Answer: The derivative of the function \(y = \ln(\cos^2x)\) with respect to x is \(\frac{dy}{dx} = -2\sin(x)\).

Step 1: Identify the Chain Rule Components

In this step, we will identify the two functions that are involved. Let's set \(u=\cos^2x\), then our given function becomes, $$y=\ln(u)$$

Step 2: Derivative of Outer Function

Now we find the derivative of outer function (\(y=\ln(u)\)) with respect to \(u\): $$\frac{dy}{du}=\frac{1}{u}$$

Step 3: Derivative of Inner Function

Now we find the derivative of inner function (\(u=\cos^2x\)) with respect to \(x\): $$\frac{du}{dx}=2\cos(x)\left(-\sin(x)\right)=-2\cos(x)\sin(x)$$

Step 4: Apply Chain Rule

Now we apply the chain rule and multiply both derivatives: $$\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}=\frac{1}{u}(-2\cos(x)\sin(x))$$

Step 5: Substitute \(u\)

Now substitute \(u\) back into the expression: $$\frac{dy}{dx}=\frac{1}{\cos^2x}(-2\cos(x)\sin(x))$$

Step 6: Simplify the Expression

Finally, we simplify the expression: $$\frac{dy}{dx}=-2\sin(x)$$ So the derivative of the given function is: $$\frac{d}{dx}\left(\ln(\cos^2x)\right)=-2\sin(x)$$