find the derivative. f(w) = (-4w^4 + 3)(csc 2w^2) a. f(w) = 128w^4 csc 2w^2 cot 2w^2 b. f(w) = (-4w csc 2w^2…

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y'=u'v + uv'$. Let $u=-4w^{4}+3$ and $v = \csc(2w^{2})$.

Step2: Find $u'$

Differentiate $u=-4w^{4}+3$ with respect to $w$. Using the power - rule $\frac{d}{dw}(aw^{n})=naw^{n - 1}$, we have $u'=-16w^{3}$.

Step3: Find $v'$

Differentiate $v=\csc(2w^{2})$ with respect to $w$. Let $t = 2w^{2}$, then $v=\csc(t)$. First, $\frac{dv}{dt}=-\csc(t)\cot(t)$ and $\frac{dt}{dw}=4w$. By the chain - rule $\frac{dv}{dw}=\frac{dv}{dt}\cdot\frac{dt}{dw}$, so $v'=-4w\csc(2w^{2})\cot(2w^{2})$.

Step4: Calculate $f'(w)$

Using the product - rule $f'(w)=u'v+uv'$. [ \begin{align*} f'(w)&=(-16w^{3})\csc(2w^{2})+(-4w^{4}+3)(-4w\csc(2w^{2})\cot(2w^{2}))\ &=-16w^{3}\csc(2w^{2})+(16w^{5}-12w)\csc(2w^{2})\cot(2w^{2})\ &=4w\csc(2w^{2})(4w^{4}\cot(2w^{2})-3\cot(2w^{2}) - 4w^{2}) \end{align*} ]

Answer:

D. $f'(w)=(4w\csc(2w^{2}))(4w^{4}\cot(2w^{2})-3\cot(2w^{2}) - 4w^{2})$