Question:

Step 1: Differentiate $x^{2x}$ with respect to $x$. We rewrite $x^{2x}$ as $e^{2x \ln x}$ to make differentiation easier. Using the chain rule: $$\frac{d}{dx} (x^{2x}) = e^{2x \ln x} \cdot \frac{d}{dx} (2x \ln x)$$ Applying the product rule to $2x \ln x$: $$\frac{d}{dx} (2x \ln x) = 2 \ln x + 2x \cdot \frac{1}{x} = 2 \ln x + 2$$ Therefore: $$\frac{d}{dx} (x^{2x}) = x^{2x} (2 \ln x + 2)$$ Step 2: Differentiate $2x^x$ with respect to $x$. We rewrite $x^x$ as $e^{x \ln x}$. Using the chain rule: $$\frac{d}{dx} (2x^x) = 2e^{x \ln x} \cdot \frac{d}{dx} (x \ln x)$$ Applying the product rule to $x \ln x$: $$\frac{d}{dx} (x \ln x) = \ln x + x \cdot \frac{1}{x} = \ln x + 1$$ Therefore: $$\frac{d}{dx} (2x^x) = 2x^x (\ln x + 1)$$ Step 3: Differentiate the term $2x^x \cot y$ with respect to $x$. Using the product rule, where $y$ is a function of $x$: $$\frac{d}{dx} (2x^x \cot y) = 2x^x (\ln x + 1) \cot y + 2x^x \cdot (-\csc^2 y) \cdot \frac{dy}{dx}$$ Step 4: Differentiate the entire implicit equation with respect to $x$. Starting with $x^{2x} - 2x^x \cot y - 1 = 0$, we differentiate both sides: $$x^{2x} (2 \ln x + 2) - 2x^x (\ln x + 1) \cot y - 2x^x \csc^2 y \cdot \frac{dy}{dx} = 0$$ Step 5: Find the value of $\cot y$ at $x = 1$. Substitute $x = 1$ into the original equation: $$1^{2 \cdot 1} - 2 \cdot 1^1 \cot y - 1 = 0$$ $$1 - 2 \cot y - 1 = 0$$ $$\cot y = 0$$ This means $y = \frac{\pi}{2}$ at $x = 1$. Step 6: Evaluate the differentiated equation at $x = 1$. Substitute $x = 1$, $\cot y = 0$, and $\csc^2\left(\frac{\pi}{2}\right) = 1$ into the differentiated equation: $$(1)^{2 \cdot 1} (2 \ln 1 + 2) - 2(1)^1 (\ln 1 + 1) \cdot 0 - 2(1)^1 (1) \cdot \frac{dy}{dx} = 0$$ Since $\ln 1 = 0$: $$1 \cdot (0 + 2) - 0 - 2 \cdot \frac{dy}{dx} = 0$$ $$2 - 2 \cdot \frac{dy}{dx} = 0$$ Step 7: Solve for $y'(1)$. $$2 = 2 \cdot \frac{dy}{dx}$$ $$\frac{dy}{dx} = 1$$ Therefore, $y'(1) = 1$, which corresponds to **Option 4**.