assume x and y are functions of t. evaluate $\frac{dy}{dt}$ for $4xy - 3x+2y^{3}=-38$, with the conditions…

Explanation:

Step1: Differentiate both sides with respect to t

Using the product - rule $(uv)^\prime = u^\prime v+uv^\prime$ and chain - rule, the derivative of $4xy$ with respect to $t$ is $4\left(x\frac{dy}{dt}+y\frac{dx}{dt}\right)$, the derivative of $-3x$ with respect to $t$ is $-3\frac{dx}{dt}$, and the derivative of $(2y)^{3}=8y^{3}$ with respect to $t$ is $24y^{2}\frac{dy}{dt}$. The derivative of the right - hand side $-38$ with respect to $t$ is $0$. So we have: $4\left(x\frac{dy}{dt}+y\frac{dx}{dt}\right)-3\frac{dx}{dt}+24y^{2}\frac{dy}{dt}=0$.

Step2: Expand the left - hand side

$4x\frac{dy}{dt}+4y\frac{dx}{dt}-3\frac{dx}{dt}+24y^{2}\frac{dy}{dt}=0$.

Step3: Group the terms with $\frac{dy}{dt}$ together

$(4x + 24y^{2})\frac{dy}{dt}+(4y - 3)\frac{dx}{dt}=0$.

Step4: Solve for $\frac{dy}{dt}$

$(4x + 24y^{2})\frac{dy}{dt}=-(4y - 3)\frac{dx}{dt}$, then $\frac{dy}{dt}=-\frac{4y - 3}{4x + 24y^{2}}\cdot\frac{dx}{dt}$.

Step5: Substitute the given values $x = 2$, $y=-2$, and $\frac{dx}{dt}=-8$

First, calculate the numerator: $4y-3=4\times(-2)-3=-8 - 3=-11$. Then, calculate the denominator: $4x + 24y^{2}=4\times2+24\times(-2)^{2}=8 + 24\times4=8 + 96 = 104$. Substitute into the formula for $\frac{dy}{dt}$: $\frac{dy}{dt}=-\frac{-11}{104}\times(-8)=-\frac{11\times8}{104}=-\frac{88}{104}=-\frac{11}{13}$.

Answer:

$-\frac{11}{13}$