Question

find the slope of the tangent line to the curve below at the point (8, 4).
\sqrt{3x + 3y}+\sqrt{3xy}=15.797958971133
slope =

Explanation:

Step1: Differentiate both sides implicitly

Differentiate $\sqrt{3x + 3y}+\sqrt{3xy}=15.797958971133$ with respect to $x$.
For $\sqrt{3x + 3y}$, using the chain - rule, we have $\frac{1}{2\sqrt{3x + 3y}}\cdot(3 + 3y')$.
For $\sqrt{3xy}$, using the product - rule and chain - rule:
Let $u = 3xy$, then $\frac{d}{dx}(\sqrt{u})=\frac{1}{2\sqrt{u}}\cdot\frac{du}{dx}$. And $\frac{du}{dx}=3y+3xy'$. So $\frac{d}{dx}(\sqrt{3xy})=\frac{3y + 3xy'}{2\sqrt{3xy}}$.
The derivative of the right - hand side is $0$.
So $\frac{3 + 3y'}{2\sqrt{3x + 3y}}+\frac{3y+3xy'}{2\sqrt{3xy}} = 0$.

Step2: Substitute $x = 8$ and $y = 4$

Substitute $x = 8$ and $y = 4$ into $\frac{3 + 3y'}{2\sqrt{3x + 3y}}+\frac{3y+3xy'}{2\sqrt{3xy}} = 0$.
First, calculate the denominators:
$\sqrt{3x + 3y}=\sqrt{3\times8 + 3\times4}=\sqrt{24 + 12}=\sqrt{36}=6$.
$\sqrt{3xy}=\sqrt{3\times8\times4}=\sqrt{96}=4\sqrt{6}$.
The equation becomes $\frac{3 + 3y'}{2\times6}+\frac{3\times4+3\times8y'}{2\times4\sqrt{6}} = 0$.
Simplify to $\frac{3 + 3y'}{12}+\frac{12 + 24y'}{8\sqrt{6}} = 0$.
Multiply through by $24\sqrt{6}$ to clear the fractions:
$2\sqrt{6}(3 + 3y')+3(12 + 24y') = 0$.
Expand: $6\sqrt{6}+6\sqrt{6}y'+36 + 72y' = 0$.
Group the terms with $y'$: $(6\sqrt{6}+72)y'=-36 - 6\sqrt{6}$.
Solve for $y'$:
\[

$$\begin{align*} y'&=\frac{-36 - 6\sqrt{6}}{6\sqrt{6}+72}\\ &=\frac{-6-\sqrt{6}}{\sqrt{6}+12}\\ &=\frac{-(6 + \sqrt{6})}{\sqrt{6}+12}\\ &=\frac{-(6+\sqrt{6})(12 - \sqrt{6})}{(12+\sqrt{6})(12 - \sqrt{6})}\\ &=\frac{-(72-6\sqrt{6}+12\sqrt{6}-6)}{144 - 6}\\ &=\frac{-(66 + 6\sqrt{6})}{138}\\ &=\frac{-11-\sqrt{6}}{23} \end{align*}$$

\]

Answer:

$\frac{-11-\sqrt{6}}{23}$