Use implicit differentiation to find $y^{'}$ and then evaluate $y^{'}$ at $(- 2, 1)$ .
$x^{3} + 8 y^{2} = \ln y$
$\begin{array}{l} y^{'} = ◻ \\ {y^{'} |}_{(- 2, 1)} = ◻ (Simplify your answer.) \end{array}$

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Answer

So, the final answer is $y^{'} = \frac{4}{5}$ .

Steps

Step 1 ：First, we differentiate both sides of the equation with respect to $x$ .

Step 2 ：The derivative of $x^{3}$ with respect to $x$ is $3 x^{2}$ .

Step 3 ：The derivative of $8 y^{2}$ with respect to $x$ is $16 y y^{'}$ .

Step 4 ：The derivative of $\ln y$ with respect to $x$ is $\frac{1}{y} y^{'}$ .

Step 5 ：So, we have $3 x^{2} + 16 y y^{'} = \frac{1}{y} y^{'}$ .

Step 6 ：Rearrange the equation to solve for $y^{'}$ , we get $y^{'} = \frac{3 x^{2} y}{16 y^{2} - 1}$ .

Step 7 ：Now, we substitute $(- 2, 1)$ into the equation, we get $y^{'} = \frac{3 (- 2)^{2} (1)}{16 (1)^{2} - 1}$ .

Step 8 ：Simplify the equation, we get $y^{'} = \frac{12}{15}$ .

Step 9 ：So, the final answer is $y^{'} = \frac{4}{5}$ .