3. - / 1 points differentiate. y = 4x^2 cos(x) cot(x) y =

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = uvw$, then $y'=u'vw + uv'w+uvw'$. Let $u = 4x^{2}$, $v=\cos(x)$, and $w = \cot(x)$.

Step2: Find $u'$

Differentiate $u = 4x^{2}$ using the power - rule. If $u = ax^{n}$, then $u'=nax^{n - 1}$. So, $u'=8x$.

Step3: Find $v'$

Differentiate $v=\cos(x)$. The derivative of $\cos(x)$ is $v'=-\sin(x)$.

Step4: Find $w'$

Differentiate $w = \cot(x)=\frac{\cos(x)}{\sin(x)}$. Using the quotient - rule $\left(\frac{f}{g}\right)'=\frac{f'g - fg'}{g^{2}}$, where $f=\cos(x)$ and $g = \sin(x)$. Then $f'=-\sin(x)$ and $g'=\cos(x)$. So, $w'=\frac{-\sin(x)\sin(x)-\cos(x)\cos(x)}{\sin^{2}(x)}=-\csc^{2}(x)$.

Step5: Calculate $y'$

$y'=u'vw+uv'w + uvw'$. Substitute $u = 4x^{2}$, $u'=8x$, $v=\cos(x)$, $v'=-\sin(x)$, $w = \cot(x)$, and $w'=-\csc^{2}(x)$ into the formula: [ \begin{align*} y'&=8x\cos(x)\cot(x)+4x^{2}(-\sin(x))\cot(x)+4x^{2}\cos(x)(-\csc^{2}(x))\ &=8x\cos(x)\cot(x)- 4x^{2}\sin(x)\cot(x)-4x^{2}\cos(x)\csc^{2}(x)\ &=8x\cos(x)\frac{\cos(x)}{\sin(x)}-4x^{2}\sin(x)\frac{\cos(x)}{\sin(x)}-4x^{2}\cos(x)\frac{1}{\sin^{2}(x)}\ &=\frac{8x\cos^{2}(x)}{\sin(x)}-4x^{2}\cos(x)-\frac{4x^{2}\cos(x)}{\sin^{2}(x)} \end{align*} ]

Answer:

$8x\cos(x)\cot(x)-4x^{2}\sin(x)\cot(x)-4x^{2}\cos(x)\csc^{2}(x)$