Questions: Find the slope of the tangent line to the curve below at the point (6,5). sqrt(3x+y) + sqrt(2xy) = 12.541798215728 slope =

Solution Steps

To find the slope of the tangent line to the curve at a given point using implicit differentiation, follow these steps:

1. Differentiate both sides of the given equation with respect to \( x \), treating \( y \) as a function of \( x \) (i.e., use the chain rule for terms involving \( y \)).
2. Solve for \( \frac{dy}{dx} \) to find the slope of the tangent line.
3. Substitute the given point \((6, 5)\) into the derivative to find the specific slope at that point.

We start with the equation of the curve given by: \[ \sqrt{3x + y} + \sqrt{2xy} = 12.541798215728 \]

Differentiating both sides with respect to \( x \) gives: \[ \frac{d}{dx} \left( \sqrt{3x + y} \right) + \frac{d}{dx} \left( \sqrt{2xy} \right) = 0 \] Using the chain rule, we find: \[ \frac{1}{2\sqrt{3x + y}} \left( 3 + \frac{dy}{dx} \right) + \frac{1}{2\sqrt{2xy}} \left( 2y + 2x\frac{dy}{dx} \right) = 0 \]

Rearranging the equation leads to: \[ \frac{dy}{dx} \left( \frac{1}{2\sqrt{3x + y}} + \frac{xy}{\sqrt{2xy}} \right) = -\left( \frac{3}{2\sqrt{3x + y}} + \frac{y}{\sqrt{2xy}} \right) \] Thus, we can express \(\frac{dy}{dx}\) as: \[ \frac{dy}{dx} = \frac{-\left( \frac{3}{2\sqrt{3x + y}} + \frac{y}{\sqrt{2xy}} \right)}{\frac{1}{2\sqrt{3x + y}} + \frac{xy}{\sqrt{2xy}}} \]

Substituting \( x = 6 \) and \( y = 5 \) into the expression for \(\frac{dy}{dx}\): \[ \frac{dy}{dx} = \frac{-\left( \frac{3 \cdot 6}{2\sqrt{3 \cdot 6 + 5}} + \frac{5}{\sqrt{2 \cdot 6 \cdot 5}} \right)}{\frac{1}{2\sqrt{3 \cdot 6 + 5}} + \frac{6 \cdot 5}{\sqrt{2 \cdot 6 \cdot 5}}} \] This simplifies to: \[ \frac{dy}{dx} = \frac{-\left( -230\sqrt{15} - 90\sqrt{23} \right)}{6\left( 5\sqrt{23} + 46\sqrt{15} \right)} \]

Final Answer