Question

use implicit differentiation to find y’ and then evaluate y’ at (1, - 2).
5xy + y + 12 = 0
y’=
y’|_{(1, - 2)}=(simplify your answer.)

Explanation:

Step1: Differentiate each term

Differentiate $5xy + y+12 = 0$ with respect to $x$.
Using the product - rule $(uv)^\prime=u^\prime v + uv^\prime$ for the term $5xy$ where $u = 5x$ and $v = y$. The derivative of $5xy$ is $5y+5xy^\prime$, the derivative of $y$ with respect to $x$ is $y^\prime$, and the derivative of the constant $12$ is $0$. So we have:
$5y + 5xy^\prime+y^\prime+0 = 0$.

Step2: Solve for $y^\prime$

Group the terms with $y^\prime$:
$5xy^\prime+y^\prime=- 5y$.
Factor out $y^\prime$:
$y^\prime(5x + 1)=-5y$.
Then $y^\prime=\frac{-5y}{5x + 1}$.

Step3: Evaluate $y^\prime$ at $(1,-2)$

Substitute $x = 1$ and $y=-2$ into $y^\prime$:
$y^\prime|_{(1,-2)}=\frac{-5\times(-2)}{5\times1 + 1}=\frac{10}{6}=\frac{5}{3}$.

Answer:

$y^\prime=\frac{-5y}{5x + 1}$
$y^\prime|_{(1,-2)}=\frac{5}{3}$