Question

use implicit differentiation to find the equation of the tangent line to the curve $xy^{3}+xy = 18$ at the point $(9,1)$. the equation of this tangent line can be written in the form $y=mx + b$ where $m$ is: and where $b$ is:

Explanation:

Step1: Differentiate both sides

Differentiate $xy^{3}+xy = 18$ with respect to $x$ using product - rule $(uv)^\prime=u^\prime v + uv^\prime$.
For $xy^{3}$, let $u = x$ and $v = y^{3}$. Then $(xy^{3})^\prime=y^{3}+3xy^{2}y^\prime$. For $xy$, let $u = x$ and $v = y$, then $(xy)^\prime=y+xy^\prime$. The derivative of the right - hand side is $0$. So $y^{3}+3xy^{2}y^\prime + y+xy^\prime=0$.

Step2: Solve for $y^\prime$

Group the terms with $y^\prime$: $3xy^{2}y^\prime+xy^\prime=-y - y^{3}$. Factor out $y^\prime$: $y^\prime(3xy^{2}+x)=-(y + y^{3})$. Then $y^\prime=-\frac{y + y^{3}}{3xy^{2}+x}$.

Step3: Find the slope $m$

Substitute $x = 9$ and $y = 1$ into $y^\prime$: $m=-\frac{1 + 1^{3}}{3\times9\times1^{2}+9}=-\frac{2}{27 + 9}=-\frac{2}{36}=-\frac{1}{18}$.

Step4: Find the y - intercept $b$

Use the point - slope form $y - y_{1}=m(x - x_{1})$ with $(x_{1},y_{1})=(9,1)$ and $m = -\frac{1}{18}$. So $y - 1=-\frac{1}{18}(x - 9)$. Expand to get $y-1=-\frac{1}{18}x+\frac{1}{2}$. Then $y=-\frac{1}{18}x+\frac{1}{2}+1=-\frac{1}{18}x+\frac{3}{2}$. So $b=\frac{3}{2}$.

Answer:

$m=-\frac{1}{18}$
$b=\frac{3}{2}$