find the gradient of the curve x^2 + 3xy +y^2 = x + 3y at point (2,-1)

To find the gradient of the curve \( x^2 + 3xy + y^2 = x + 3y \) at the point \((2, -1)\), we will use implicit differentiation.

### Step 1: Differentiate the equation with respect to \(x\).
Given the equation:

\[
x^2 + 3xy + y^2 = x + 3y
\]

Differentiate both sides with respect to \(x\), remembering to apply the product rule to \(3xy\) and the chain rule to \(y^2\) and \(3y\) (since \(y\) is a function of \(x\)):

\[
\frac{d}{dx}(x^2) + \frac{d}{dx}(3xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(x) + \frac{d}{dx}(3y)
\]

This gives us:

\[
2x + 3\left(\frac{dy}{dx} \cdot x + y \cdot \frac{d}{dx}(x)\right) + 2y\frac{dy}{dx} = 1 + 3\frac{dy}{dx}
\]

Simplifying:

\[
2x + 3\left(x\frac{dy}{dx} + y\right) + 2y\frac{dy}{dx} = 1 + 3\frac{dy}{dx}
\]

Expanding and rearranging the terms to collect \(\frac{dy}{dx}\) on one side:

\[
2x + 3x\frac{dy}{dx} + 3y + 2y\frac{dy}{dx} = 1 + 3\frac{dy}{dx}
\]

\[
(3x + 2y)\frac{dy}{dx} - 3\frac{dy}{dx} = 1 - 2x - 3y
\]

Factor out \(\frac{dy}{dx}\):

\[
(3x + 2y - 3)\frac{dy}{dx} = 1 - 2x - 3y
\]

Now, solve for \(\frac{dy}{dx}\):

\[
\frac{dy}{dx} = \frac{1 - 2x - 3y}{3x + 2y - 3}
\]

### Step 2: Substitute the point \((2, -1)\) into the derivative.

Substitute \(x = 2\) and \(y = -1\):

\[
\frac{dy}{dx} = \frac{1 - 2(2) - 3(-1)}{3(2) + 2(-1) - 3}
\]

Simplify the expression:

\[
\frac{dy}{dx} = \frac{1 - 4 + 3}{6 - 2 - 3}
\]

\[
\frac{dy}{dx} = \frac{0}{1} = 0
\]

### Conclusion:

The gradient of the curve at the point \((2, -1)\) is \( \boxed{0} \).

Would you like more details, or do you have any further questions?

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Here are some related questions you might consider:

1. What is implicit differentiation, and when is it used?
2. How does the product rule work in differentiation?
3. What is the geometric interpretation of a gradient being zero?
4. How would the gradient change if the point on the curve were different?
5. How do you find the equation of the tangent line at the point \((2, -1)\)?
6. What happens to the curve near the point \((2, -1)\) given the gradient?
7. Can you find the second derivative of the curve at the point \((2, -1)\)?
8. How would this problem be different if the curve were parametrized?

**Tip:** Always double-check your differentiation steps, especially when dealing with implicit differentiation, as errors can easily occur in handling the terms.

Learn how to find the gradient of the curve x^2 + 3xy + y^2 = x + 3y at the specific point (2, -1) using implicit differentiation. This detailed solution covers the application of implicit differentiation, product rule, and chain rule in calculus. Suitable for advanced high school students and beyond.

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