Question

1. - / 5.26 points find the derivative of the function. y = cot²(cos θ) y = i

Explanation:

Step1: Apply chain rule to outer function

Let \( u = \cot(\cos\theta) \), so \( y = u^2 \). Derivative: \( y' = 2u \cdot u' = 2\cot(\cos\theta) \cdot \frac{d}{d\theta}[\cot(\cos\theta)] \)

Step2: Differentiate inner cot function

Derivative of \( \cot(v) \) is \( -\csc^2(v) \cdot v' \). Let \( v = \cos\theta \), so \( \frac{d}{d\theta}[\cot(\cos\theta)] = -\csc^2(\cos\theta) \cdot \frac{d}{d\theta}[\cos\theta] \)

Step3: Differentiate innermost cosine

Derivative of \( \cos\theta \) is \( -\sin\theta \)

Step4: Combine all parts

Substitute back: \( y' = 2\cot(\cos\theta) \cdot [-\csc^2(\cos\theta) \cdot (-\sin\theta)] = 2\sin\theta \cot(\cos\theta) \csc^2(\cos\theta) \)

Answer:

\( 2\sin\theta \cot(\cos\theta) \csc^2(\cos\theta) \)