Question

1. - / 5.55 points differentiate. $y = 8x^2 \cos x \cot x$ $y = $

Explanation:

Step1: Apply product rule to three functions

Let \( u=8x^2 \), \( v=\cos x \), \( w=\cot x \). Product rule: \( y'=u'vw + uv'w + uvw' \)

Step2: Compute derivatives of u, v, w

\( u'=16x \), \( v'=-\sin x \), \( w'=-\csc^2 x \)

Step3: Substitute into product rule

\( y'=16x \cdot \cos x \cdot \cot x + 8x^2 \cdot (-\sin x) \cdot \cot x + 8x^2 \cdot \cos x \cdot (-\csc^2 x) \)

Step4: Simplify terms

\( \cot x = \frac{\cos x}{\sin x} \), so \( -\sin x \cot x = -\cos x \); \( \csc^2 x = \frac{1}{\sin^2 x} \)
\( y' = 16x\cos x\cot x - 8x^2\cos x - 8x^2\cos x\csc^2 x \)

Answer:

\( 16x\cos x\cot x - 8x^2\cos x - 8x^2\cos x\csc^2 x \)