Question

questions

1. the concentration of a dye (in ml) in a solution is modeled by

c(t)=\frac{50ln(1 + 2t^{2})}{t^{3}+2t + 3},tgeq0\text{ minutes.}
find the rate of change of concentration at (t = 4) minutes. include units.

1. differentiate the following function:

f(x)=left(7x^{5}+\frac{3}{sqrt{x}}
ight)^{2}e^{x^{3}+2x}

Explanation:

Step1: Recall quotient - rule for differentiation

If $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = 50\ln(1 + 2t^{2})$ and $v=t^{3}+2t + 3$.
First, find $u^\prime$ and $v^\prime$.
For $u = 50\ln(1 + 2t^{2})$, by the chain - rule, if $y=\ln(u)$ and $u = 1+2t^{2}$, then $\frac{dy}{du}=\frac{1}{u}$ and $\frac{du}{dt}=4t$. So $u^\prime=50\times\frac{4t}{1 + 2t^{2}}=\frac{200t}{1 + 2t^{2}}$.
For $v=t^{3}+2t + 3$, $v^\prime=3t^{2}+2$.

Step2: Apply the quotient - rule

$C^\prime(t)=\frac{\frac{200t}{1 + 2t^{2}}(t^{3}+2t + 3)-50\ln(1 + 2t^{2})(3t^{2}+2)}{(t^{3}+2t + 3)^{2}}$.

Step3: Substitute $t = 4$

First, calculate the values:
When $t = 4$, $1+2t^{2}=1 + 2\times16=33$, $t^{3}+2t + 3=64 + 8+3=75$, $3t^{2}+2=3\times16 + 2=50$.
$u^\prime\big|_{t = 4}=\frac{200\times4}{33}=\frac{800}{33}$, $v^\prime\big|_{t = 4}=50$.
$C^\prime(4)=\frac{\frac{800}{33}\times75-50\ln(33)\times50}{75^{2}}$
$=\frac{\frac{60000}{33}-2500\ln(33)}{5625}$
$=\frac{60000 - 82500\ln(33)}{33\times5625}$
$=\frac{60000 - 82500\ln(33)}{185625}\text{ mL/min}$.

for second question:

Step1: Recall the product - rule and chain - rule

If $y = uv$, then $y^\prime=u^\prime v+uv^\prime$. Here, $u=(7x^{5}+\frac{3}{\sqrt{x}})^{2}$ and $v = e^{x^{3}+2x}$.
First, use the chain - rule to find $u^\prime$. Let $w = 7x^{5}+3x^{-\frac{1}{2}}$, then $u = w^{2}$. So $u^\prime = 2w\times w^\prime$.
$w^\prime=35x^{4}-\frac{3}{2}x^{-\frac{3}{2}}$. Then $u^\prime=2(7x^{5}+\frac{3}{\sqrt{x}})(35x^{4}-\frac{3}{2\sqrt{x^{3}}})$.
For $v = e^{x^{3}+2x}$, by the chain - rule, if $y = e^{u}$ and $u=x^{3}+2x$, then $v^\prime=(3x^{2}+2)e^{x^{3}+2x}$.

Step2: Apply the product - rule

$f^\prime(x)=u^\prime v+uv^\prime$
$=2(7x^{5}+\frac{3}{\sqrt{x}})(35x^{4}-\frac{3}{2\sqrt{x^{3}}})e^{x^{3}+2x}+(7x^{5}+\frac{3}{\sqrt{x}})^{2}(3x^{2}+2)e^{x^{3}+2x}$
$=(7x^{5}+\frac{3}{\sqrt{x}})e^{x^{3}+2x}(2(35x^{4}-\frac{3}{2\sqrt{x^{3}}})+(7x^{5}+\frac{3}{\sqrt{x}})(3x^{2}+2))$.

Answer:

$\frac{60000 - 82500\ln(33)}{185625}\text{ mL/min}$