use implicit differentiation to find y’ and then evaluate y’ at (2,6). 8xy + y - 102 = 0 y’ = y’|(2,6) =…

Explanation:

Step1: Differentiate both sides

Differentiate $8xy + y-102 = 0$ with respect to $x$. Using the product - rule $(uv)^\prime=u^\prime v + uv^\prime$ for $8xy$ where $u = 8x$ and $v = y$. The derivative of $8xy$ is $8y+8xy^\prime$, the derivative of $y$ is $y^\prime$ and the derivative of $102$ is $0$. So, $\frac{d}{dx}(8xy)+\frac{d}{dx}(y)-\frac{d}{dx}(102)=\frac{d}{dx}(0)$ gives $8y + 8xy^\prime+y^\prime-0 = 0$.

Step2: Solve for $y^\prime$

Group the terms with $y^\prime$ together: $8xy^\prime+y^\prime=- 8y$. Factor out $y^\prime$: $y^\prime(8x + 1)=-8y$. Then $y^\prime=\frac{-8y}{8x + 1}$.

Step3: Evaluate $y^\prime$ at $(2,6)$

Substitute $x = 2$ and $y = 6$ into $y^\prime$. $y^\prime|_{(2,6)}=\frac{-8\times6}{8\times2+1}=\frac{-48}{16 + 1}=\frac{-48}{17}$.

Answer:

$y^\prime=\frac{-8y}{8x + 1}$ $y^\prime|_{(2,6)}=-\frac{48}{17}$