instructions\n• work this problem out on 1 sheet of paper, one - side only. label the problem, \2 parts.\\n•…

Explanation:

Step1: Recall quotient - rule

The quotient rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. For $y = \frac{x^{5}}{3x^{7}-6}$, let $u = x^{5}$ and $v=3x^{7}-6$. First, find $u^\prime$ and $v^\prime$. $u^\prime=\frac{d}{dx}(x^{5}) = 5x^{4}$ and $v^\prime=\frac{d}{dx}(3x^{7}-6)=21x^{6}$.

Step2: Apply quotient - rule

$y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}=\frac{5x^{4}(3x^{7}-6)-x^{5}(21x^{6})}{(3x^{7}-6)^{2}}$. Expand the numerator: [ \begin{align*} 5x^{4}(3x^{7}-6)-x^{5}(21x^{6})&=15x^{11}-30x^{4}-21x^{11}\ &=- 6x^{11}-30x^{4} \end{align*} ] So, $\frac{d}{dx}\left[\frac{x^{5}}{3x^{7}-6}\right]=\frac{-6x^{11}-30x^{4}}{(3x^{7}-6)^{2}}$.

Step3: Recall chain - rule for part b

The chain - rule states that if $y = f(g(x))$, then $y^\prime=f^\prime(g(x))\cdot g^\prime(x)$. For $y=\sec(\cos(\pi x))$, let $u = \cos(\pi x)$ and $y=\sec(u)$. First, find $\frac{dy}{du}$ and $\frac{du}{dx}$. $\frac{dy}{du}=\sec(u)\tan(u)$ and $\frac{du}{dx}=-\pi\sin(\pi x)$.

Step4: Apply chain - rule

$\frac{d}{dx}[\sec(\cos(\pi x))]=\sec(\cos(\pi x))\tan(\cos(\pi x))\cdot(-\pi\sin(\pi x))=-\pi\sin(\pi x)\sec(\cos(\pi x))\tan(\cos(\pi x))$.

Answer:

a. $\frac{-6x^{11}-30x^{4}}{(3x^{7}-6)^{2}}$ b. $-\pi\sin(\pi x)\sec(\cos(\pi x))\tan(\cos(\pi x))$