If `2x^(y) +3y^(x) =20`, then `(dy)/(dx)` at (2,2) is equal to :

Options

A. `-((2+log_e8)/(3+log_e4))`
B. `-((3+log_e16)/(4+log_e8))`
C. `-((3+log_e8)/(2+log_e4))`
D. `-((3+log_e4)/(2+log_e8))`

Answer

To solve the given equation \(2x^{y} + 3y^{x} = 20\) and find \(\frac{dy}{dx}\) at the point (2,2), we will use implicit differentiation.

### Step-by-Step Solution:

1. **Differentiate the equation implicitly**:
We start with the equation:
\[ 2x^{y} + 3y^{x} = 20 \]
We will differentiate both sides with respect to \(x\).

2. **Differentiate the first term \(2x^{y}\)**:
Using the product rule:
\[ \frac{d}{dx}(2x^{y}) = 2 \left( y \cdot x^{y-1} \cdot \frac{dx}{dx} + x^{y} \cdot \ln(x) \cdot \frac{dy}{dx} \right) = 2 \left( yx^{y-1} + x^{y} \ln(x) \frac{dy}{dx} \right) \]

3. **Differentiate the second term \(3y^{x}\)**:
Again using the product rule:
\[ \frac{d}{dx}(3y^{x}) = 3 \left( y^{x} \cdot \ln(y) \cdot \frac{dy}{dx} + x \cdot y^{x-1} \cdot \frac{dy}{dx} \right) = 3 \left( y^{x} \ln(y) \frac{dy}{dx} + xy^{x-1} \frac{dy}{dx} \right) \]

4. **Set the derivative equal to zero**:
Since the right side of the original equation is a constant (20), its derivative is 0:
\[ 2 \left( yx^{y-1} + x^{y} \ln(x) \frac{dy}{dx} \right) + 3 \left( y^{x} \ln(y) \frac{dy}{dx} + xy^{x-1} \frac{dy}{dx} \right) = 0 \]

5. **Substitute the point (2,2)**:
Now we substitute \(x = 2\) and \(y = 2\):
\[ 2 \left( 2 \cdot 2^{1} + 2^{2} \ln(2) \frac{dy}{dx} \right) + 3 \left( 2^{2} \ln(2) \frac{dy}{dx} + 2 \cdot 2^{1} \frac{dy}{dx} \right) = 0 \]
Simplifying:
\[ 2(4 + 4 \ln(2) \frac{dy}{dx}) + 3(4 \ln(2) \frac{dy}{dx} + 4 \frac{dy}{dx}) = 0 \]
\[ 8 + 8 \ln(2) \frac{dy}{dx} + 12 \ln(2) \frac{dy}{dx} + 12 \frac{dy}{dx} = 0 \]

6. **Combine like terms**:
\[ 8 + (8 \ln(2) + 12 \ln(2) + 12) \frac{dy}{dx} = 0 \]
\[ 8 + (20 \ln(2) + 12) \frac{dy}{dx} = 0 \]

7. **Solve for \(\frac{dy}{dx}\)**:
\[ (20 \ln(2) + 12) \frac{dy}{dx} = -8 \]
\[ \frac{dy}{dx} = \frac{-8}{20 \ln(2) + 12} \]

### Final Answer:
Thus, at the point (2,2), the value of \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = \frac{-8}{20 \ln(2) + 12} \]