Answer:
Step-To find the slope of the tangent line to the curve defined by the equation \(3x + 3y + 2xy = 15.189269908398\), you can follow these steps:
1. Differentiate the equation with respect to \(x\) implicitly to find the derivative \(\frac{dy}{dx}\).
\(3x + 3y + 2xy = 15.189269908398\)
Now, differentiate both sides with respect to \(x\):
\(\frac{d}{dx}(3x) + \frac{d}{dx}(3y) + \frac{d}{dx}(2xy) = \frac{d}{dx}(15.189269908398)\)
2. Solve for \(\frac{dy}{dx}\):
The derivative of \(3x\) with respect to \(x\) is simply \(3\), and the derivative of a constant (like \(15.189269908398\)) is zero. For the middle term, you need to apply the product rule:
\(\frac{d}{dx}(3y) = 3\frac{dy}{dx}\)
Now, for the last term, \(\frac{d}{dx}(2xy)\), you can apply the product rule as well:
\(\frac{d}{dx}(2xy) = 2x\frac{dy}{dx} + 2y\frac{dx}{dx}\)
Since \(\frac{dx}{dx}\) is 1, this simplifies to:
\(2x\frac{dy}{dx} + 2y\)
Now, you have:
\(3 + 3\frac{dy}{dx} + 2x\frac{dy}{dx} + 2y = 0\)
3. Combine like terms:
\(3\frac{dy}{dx} + 2x\frac{dy}{dx} + 2y = -3\)
4. Factor out \(\frac{dy}{dx}\):
\(\frac{dy}{dx}(3 + 2x) + 2y = -3\)
5. Isolate \(\frac{dy}{dx}\):
\(\frac{dy}{dx} = \frac{-3 - 2y}{3 + 2x}\)
6. Now, you can find the slope of the tangent line at a specific point. If you have a particular point \((x_0, y_0)\) on the curve, plug these values into the equation for \(\frac{dy}{dx}\):
\(\text{Slope} = \frac{-3 - 2y_0}{3 + 2x_0}\)
Without knowing the specific point \((x_0, y_0)\), I can't provide an exact numerical value for the slope. You'll need to substitute the values of \(x_0\) and \(y_0\) from the point where you want to find the tangent slope.by-step
Step-by-step explanation: