Let y be an implicit function of x defined by $x^{2x}-2x^{x}cot~y-1=0$. Then $y^{\prime}(1)$ equals:

Question

Accepted Answer

To determine $ y'(1) $ for the implicit function defined by the equation $ x^{2x} - 2x^x \cot y - 1 = 0 $, we need to differentiate the equation with respect to $ x $ and then evaluate the derivative at $ x = 1 $.

### Step-by-Step Explanation:

1. **Differentiate the Equation Implicitly:**
   The given equation is:
   $$
   x^{2x} - 2x^x \cot y - 1 = 0
   $$
   Differentiate both sides with respect to $ x $.

2. **Differentiate $ x^{2x} $:**
   Using the chain rule and properties of logarithms, the derivative of $ x^{2x} $ is:
   $$
   \frac{d}{dx}(x^{2x}) = x^{2x} \left(2 \ln x + 2\right)
   $$

3. **Differentiate $ -2x^x \cot y $:**
   Using the product rule and chain rule:
   $$
   \frac{d}{dx}(-2x^x \cot y) = -2 \left( x^x \ln x \cot y + x^x (-\csc^2 y) y' \right)
   $$

4. **Differentiate the Constant:**
   The derivative of $-1$ is $0$.

5. **Combine the Derivatives:**
   Combine all parts:
   $$
   x^{2x} (2 \ln x + 2) - 2 \left( x^x \ln x \cot y + x^x (-\csc^2 y) y' \right) = 0
   $$

6. **Evaluate at $ x = 1 $:**
   Substitute $ x = 1 $ into the differentiated equation:
   - $ 1^{2 \cdot 1} = 1 $
   - $ \ln 1 = 0 $
   - The equation simplifies to:
     $$
     1 \cdot (2 \cdot 0 + 2) - 2 \left( 1 \cdot 0 \cdot \cot y + 1 \cdot (-\csc^2 y) y' \right) = 0
     $$
   - Simplifying further:
     $$
     2 + 2 \csc^2 y \cdot y' = 0
     $$

7. **Solve for $ y'(1) $:**
   $$
   2 \csc^2 y \cdot y' = -2
   $$
   $$
   y' = -\frac{2}{2 \csc^2 y} = -\frac{1}{\csc^2 y} = -\sin^2 y
   $$

8. **Determine $ y $ at $ x = 1 $:**
   From the original equation at $ x = 1 $:
   $$
   1^{2 \cdot 1} - 2 \cdot 1^1 \cot y - 1 = 0 \implies -2 \cot y = 0 \implies \cot y = 0
   $$
   This implies $ y = \frac{\pi}{2} $ (since $\cot \frac{\pi}{2} = 0$).

9. **Calculate $ y'(1) $:**
   $$
   y' = -\sin^2 \left(\frac{\pi}{2}\right) = -1
   $$

### Conclusion:

- **Option 1 (-1) is correct** because the calculation shows that $ y'(1) = -1 $.

### Explanation for Incorrect Options:

- **Option 2 (1):** This is incorrect because the derivative calculation results in $-1$, not $1$.
- **Option 3 ($\log 2$):** This is incorrect because the derivative does not involve any logarithmic terms that would result in $\log 2$.
- **Option 4 ($-\log 2$):** This is incorrect for the same reason as Option 3; the derivative does not yield $-\log 2$.

Thus, the correct answer is indeed Option 1: $-1$.