A curve has the equation 2^x + y^2 = 2xy. Find the exact value of dy/dx at the point on the function with coordinates (3,2)

To find the exact value of \(\frac{dy}{dx}\) at the point \((3, 2)\) for the curve given by the equation \(2^x + y^2 = 2xy\), we follow these steps:

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

Given equation:
\[
2^x + y^2 = 2xy
\]

Differentiate both sides with respect to \(x\):

- The derivative of \(2^x\) with respect to \(x\) is \(2^x \ln(2)\).
- The derivative of \(y^2\) with respect to \(x\) is \(2y \frac{dy}{dx}\) (using the chain rule).
- The derivative of \(2xy\) with respect to \(x\) involves the product rule:
  \[
  \frac{d}{dx}(2xy) = 2 \left(\frac{dy}{dx} \cdot x + y \right)
  \]

Thus, differentiating the equation gives:
\[
2^x \ln(2) + 2y \frac{dy}{dx} = 2\left(\frac{dy}{dx} \cdot x + y\right)
\]

### Step 2: Solve for \(\frac{dy}{dx}\)

Rearrange the equation to isolate \(\frac{dy}{dx}\):

\[
2^x \ln(2) + 2y \frac{dy}{dx} = 2x \frac{dy}{dx} + 2y
\]

Move all terms involving \(\frac{dy}{dx}\) to one side:

\[
2y \frac{dy}{dx} - 2x \frac{dy}{dx} = 2y - 2^x \ln(2)
\]

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

\[
\frac{dy}{dx} (2y - 2x) = 2y - 2^x \ln(2)
\]

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

\[
\frac{dy}{dx} = \frac{2y - 2^x \ln(2)}{2y - 2x}
\]

### Step 3: Evaluate \(\frac{dy}{dx}\) at the point \((3, 2)\)

Substitute \(x = 3\) and \(y = 2\) into the equation:

\[
\frac{dy}{dx} = \frac{2(2) - 2^3 \ln(2)}{2(2) - 2(3)}
\]

Simplify the expression:

- \(2^3 = 8\), so the numerator becomes:
  \[
  4 - 8 \ln(2)
  \]
- The denominator becomes:
  \[
  4 - 6 = -2
  \]

Thus:
\[
\frac{dy}{dx} = \frac{4 - 8\ln(2)}{-2}
\]

Simplify further:

\[
\frac{dy}{dx} = \frac{4(1 - 2\ln(2))}{-2} = -2(1 - 2\ln(2))
\]

Thus, the exact value of \(\frac{dy}{dx}\) at the point \((3, 2)\) is:

\[
\boxed{2(2\ln(2) - 1)}
\]

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Would you like further explanation or have any additional questions?

Here are some related questions to explore:
1. How does implicit differentiation differ from explicit differentiation?
2. What is the significance of the natural logarithm in calculus problems?
3. How can you determine if a point is a critical point for a given curve?
4. What are the implications of finding a negative derivative value at a specific point?
5. How can product and chain rules be combined to solve complex differentiation problems?

**Tip:** When evaluating derivatives at specific points, always double-check the substitution process to ensure accuracy in both the numerator and denominator.

Find the exact value of dy/dx at the point (3, 2) for the curve given by the equation 2^x + y^2 = 2xy. Detailed step-by-step solution including implicit differentiation, product rule, and chain rule applications.

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