if -3x^3 + y^3 = -3 then find dy/dx at the point (-2, -3). answer dy/dx|_(-2,-3) =

Explanation:

Step1: Differentiate both sides

Differentiate $-3x^{3}+y^{3}=-3$ with respect to $x$. Using the power - rule and chain - rule, we have: $\frac{d}{dx}(-3x^{3})+\frac{d}{dx}(y^{3})=\frac{d}{dx}(-3)$. The derivative of $-3x^{3}$ with respect to $x$ is $-9x^{2}$, and for $y^{3}$ with respect to $x$ is $3y^{2}\frac{dy}{dx}$, and the derivative of a constant $-3$ is $0$. So, $-9x^{2}+3y^{2}\frac{dy}{dx}=0$.

Step2: Solve for $\frac{dy}{dx}$

Isolate $\frac{dy}{dx}$ in the equation $-9x^{2}+3y^{2}\frac{dy}{dx}=0$. First, move $-9x^{2}$ to the other side: $3y^{2}\frac{dy}{dx}=9x^{2}$. Then divide both sides by $3y^{2}$ to get $\frac{dy}{dx}=\frac{9x^{2}}{3y^{2}}=\frac{3x^{2}}{y^{2}}$.

Step3: Evaluate at the given point

Substitute $x = - 2$ and $y=-3$ into $\frac{dy}{dx}=\frac{3x^{2}}{y^{2}}$. $\frac{dy}{dx}\big|_{(-2,-3)}=\frac{3\times(-2)^{2}}{(-3)^{2}}=\frac{3\times4}{9}=\frac{4}{3}$.

Answer:

$\frac{4}{3}$