use implicit differentiation to find y’ and then evaluate y’ at (-2,1). x³ + 8y³ = ln y y’ = y’|(-2,1) =…

Explanation:

Step1: Differentiate both sides

Differentiate $x^{3}+8y^{3}=\ln y$ with respect to $x$. The derivative of $x^{3}$ with respect to $x$ is $3x^{2}$. For $8y^{3}$, using the chain - rule, we get $24y^{2}y'$. The derivative of $\ln y$ with respect to $x$ is $\frac{y'}{y}$. So, $3x^{2}+24y^{2}y'=\frac{y'}{y}$.

Step2: Isolate $y'$

First, move all terms with $y'$ to one side: $24y^{2}y'-\frac{y'}{y}=- 3x^{2}$. Factor out $y'$: $y'(24y^{2}-\frac{1}{y})=-3x^{2}$. Then $y'=\frac{-3x^{2}}{24y^{2}-\frac{1}{y}}=\frac{-3x^{2}y}{24y^{3}-1}$.

Step3: Evaluate $y'$ at $(-2,1)$

Substitute $x = - 2$ and $y = 1$ into $y'$: $y'\big|_{(-2,1)}=\frac{-3\times(-2)^{2}\times1}{24\times1^{3}-1}=\frac{-3\times4}{24 - 1}=\frac{-12}{23}$.

Answer:

$y'=\frac{-3x^{2}y}{24y^{3}-1}$ $y'\big|_{(-2,1)}=-\frac{12}{23}$