find diferentiation of (i) `x^(y)cdoty^(x)=1` (ii) `y=e^(x^(x))`

Answer

To find the differentiation of the given equations, we will solve each part step by step.

### Part (i): Differentiate \( x^y \cdot y^x = 1 \)

1. **Take the logarithm of both sides:**
\[ \log(x^y \cdot y^x) = \log(1) \]
Since \(\log(1) = 0\), we have:
\[ \log(x^y) + \log(y^x) = 0 \]

2. **Apply the logarithmic properties:**
Using the property \(\log(a^b) = b \log(a)\):
\[ y \log(x) + x \log(y) = 0 \]

3. **Differentiate both sides with respect to \(x\):**
\[ \frac{d}{dx}(y \log(x)) + \frac{d}{dx}(x \log(y)) = 0 \]
Using the product rule on both terms:
\[ \frac{dy}{dx} \log(x) + y \cdot \frac{1}{x} + \log(y) + x \cdot \frac{dy}{dx} \cdot \frac{1}{y} = 0 \]

4. **Rearranging the equation:**
Grouping the terms involving \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} \left( \log(x) + \frac{x}{y} \right) = -\left( \frac{y}{x} + \log(y) \right) \]

5. **Solve for \(\frac{dy}{dx}\):**
\[ \frac{dy}{dx} = -\frac{\frac{y}{x} + \log(y)}{\log(x) + \frac{x}{y}} \]

### Part (ii): Differentiate \( y = e^{x^x} \)

1. **Take the logarithm of both sides:**
\[ \log(y) = \log(e^{x^x}) \]
Simplifying gives:
\[ \log(y) = x^x \]

2. **Differentiate both sides with respect to \(x\):**
Using implicit differentiation:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x^x) \]
To differentiate \(x^x\), use the property:
\[ x^x = e^{x \log(x)} \]
Thus:
\[ \frac{d}{dx}(x^x) = e^{x \log(x)} \left( \log(x) + 1 \right) = x^x (\log(x) + 1) \]

3. **Substituting back:**
\[ \frac{1}{y} \frac{dy}{dx} = x^x (\log(x) + 1) \]

4. **Solve for \(\frac{dy}{dx}\):**
\[ \frac{dy}{dx} = y \cdot x^x (\log(x) + 1) \]
Since \(y = e^{x^x}\), we can write:
\[ \frac{dy}{dx} = e^{x^x} \cdot x^x (\log(x) + 1) \]

### Summary of Results:
1. For \( x^y \cdot y^x = 1 \):
\[ \frac{dy}{dx} = -\frac{\frac{y}{x} + \log(y)}{\log(x) + \frac{x}{y}} \]
2. For \( y = e^{x^x} \):
\[ \frac{dy}{dx} = e^{x^x} \cdot x^x (\log(x) + 1) \]