Question

\(\frac{d}{dx} \left[ (1 - \cos 2x) + 2\cos^2 x \right] \)

• A. \( -4 \sin x \cos x \)
• B. 1
• C. 0
• D. 2

Hint:

When differentiating trigonometric functions involving multiple angles, be sure to apply the chain rule carefully. For \( \cos^2 x \), use the identity and then differentiate.

Answer 1

The Correct Option is C

We need to differentiate the function: \[ f(x) = (1 - \cos 2x) + 2 \cos^2 x \] Step 1: Differentiate \( 1 - \cos 2x \)
The derivative of a constant is 0, so we only need to differentiate \( \cos 2x \). Using the chain rule, we get: \[ \frac{d}{dx} \cos 2x = -\sin 2x \cdot 2 = -2 \sin 2x \] Thus, the derivative of \( 1 - \cos 2x \) is \( 2 \sin 2x \).
Step 2: Differentiate \( 2 \cos^2 x \)
We use the chain rule again to differentiate \( 2 \cos^2 x \). First, differentiate the outer function: \[ \frac{d}{dx} \left( \cos^2 x \right) = 2 \cos x \cdot (-\sin x) = -2 \cos x \sin x \] Thus, the derivative of \( 2 \cos^2 x \) is \( -4 \cos x \sin x \).
Step 3: Combine the results
Now, combine the results of both derivatives: \[ \frac{d}{dx} \left[ (1 - \cos 2x) + 2\cos^2 x \right] = 2 \sin 2x - 4 \cos x \sin x \] Now notice that \( 2 \sin 2x = 4 \sin x \cos x \). Thus: \[ 4 \sin x \cos x - 4 \cos x \sin x = 0 \] So, the derivative is \( 0 \).