quiz\n4. let c(t) = 7 / (50 + 4t). find c(t) at t = 5 using quotient rule\n5. let y=(2x³ - 4e^x)^4. find dy/dx

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. For $C(t)=\frac{7}{50 + 4t}$, we have $u = 7$, $u'=0$, $v=50 + 4t$, and $v' = 4$.

Step2: Apply quotient - rule

$C'(t)=\frac{0\times(50 + 4t)-7\times4}{(50 + 4t)^{2}}=\frac{- 28}{(50 + 4t)^{2}}$.

Step3: Evaluate at $t = 5$

Substitute $t = 5$ into $C'(t)$: $C'(5)=\frac{-28}{(50+4\times5)^{2}}=\frac{-28}{(50 + 20)^{2}}=\frac{-28}{4900}=-\frac{1}{175}$.

Step4: For $y=(2x^{3}-4e^{x})^{4}$, use chain - rule

Let $u = 2x^{3}-4e^{x}$, then $y = u^{4}$. By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. First, $\frac{dy}{du}=4u^{3}=4(2x^{3}-4e^{x})^{3}$. Second, $\frac{du}{dx}=6x^{2}-4e^{x}$.

Step5: Calculate $\frac{dy}{dx}$

$\frac{dy}{dx}=4(2x^{3}-4e^{x})^{3}(6x^{2}-4e^{x})$.

Answer:

$C'(5)=-\frac{1}{175}$; $\frac{dy}{dx}=4(2x^{3}-4e^{x})^{3}(6x^{2}-4e^{x})$