Find `(dy)/(dx)`, when:
`y=x^(1//x)`

Answer

To find \(\frac{dy}{dx}\) when \(y = x^{\frac{1}{x}}\), we will follow these steps:

### Step 1: Take the natural logarithm of both sides
We start by taking the logarithm of both sides to simplify the expression:
\[ \ln y = \ln\left(x^{\frac{1}{x}}\right) \]

### Step 2: Use the property of logarithms
Using the property of logarithms that states \(\ln(a^b) = b \ln a\), we can rewrite the right-hand side:
\[ \ln y = \frac{1}{x} \ln x \]

### Step 3: Differentiate both sides
Now, we differentiate both sides with respect to \(x\). Remember to use implicit differentiation on the left side:
\[ \frac{d}{dx}(\ln y) = \frac{d}{dx}\left(\frac{1}{x} \ln x\right) \]
Using the chain rule on the left side, we have:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}\left(\frac{1}{x} \ln x\right) \]

### Step 4: Differentiate the right-hand side
To differentiate the right-hand side, we will use the product rule:
\[ \frac{d}{dx}\left(\frac{1}{x} \ln x\right) = \frac{d}{dx}(\frac{1}{x}) \cdot \ln x + \frac{1}{x} \cdot \frac{d}{dx}(\ln x) \]
Calculating each derivative:
\[ \frac{d}{dx}\left(\frac{1}{x}\right) = -\frac{1}{x^2}, \quad \frac{d}{dx}(\ln x) = \frac{1}{x} \]
Thus, we have:
\[ \frac{d}{dx}\left(\frac{1}{x} \ln x\right) = -\frac{1}{x^2} \ln x + \frac{1}{x^2} \]
Combining these:
\[ \frac{d}{dx}\left(\frac{1}{x} \ln x\right) = \frac{1 - \ln x}{x^2} \]

### Step 5: Substitute back into the equation
Now substituting back into our differentiated equation:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{1 - \ln x}{x^2} \]

### Step 6: Solve for \(\frac{dy}{dx}\)
Multiplying both sides by \(y\):
\[ \frac{dy}{dx} = y \cdot \frac{1 - \ln x}{x^2} \]
Since \(y = x^{\frac{1}{x}}\), we substitute \(y\) back:
\[ \frac{dy}{dx} = x^{\frac{1}{x}} \cdot \frac{1 - \ln x}{x^2} \]

### Final Answer
Thus, the derivative \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = \frac{x^{\frac{1}{x}}(1 - \ln x)}{x^2} \]

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