If `x = e^(x//y)` , then prove that `(dy)/(dx) = (x-y)/(xlogx)`.

Explanation

To prove that \(\frac{dy}{dx} = \frac{x - y}{x \log x}\) given \(x = e^{\frac{x}{y}}\), we will follow these steps:

### Step 1: Take the natural logarithm of both sides
Starting with the equation:
\[ x = e^{\frac{x}{y}} \]
We take the natural logarithm (ln) of both sides:
\[ \ln x = \ln\left(e^{\frac{x}{y}}\right) \]
Using the property of logarithms, we simplify the right side:
\[ \ln x = \frac{x}{y} \]

### Step 2: Rearrange the equation
From the equation \(\ln x = \frac{x}{y}\), we can rearrange it to express \(y\) in terms of \(x\):
\[ y = \frac{x}{\ln x} \]

### Step 3: Differentiate both sides with respect to \(x\)
Now we differentiate \(y\) with respect to \(x\):
\[ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{x}{\ln x}\right) \]
We will use the quotient rule for differentiation, which states that if \(u = x\) and \(v = \ln x\), then:
\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
Calculating the derivatives:
- \(\frac{du}{dx} = 1\)
- \(\frac{dv}{dx} = \frac{1}{x}\)

Substituting these into the quotient rule gives:
\[ \frac{dy}{dx} = \frac{\ln x \cdot 1 - x \cdot \frac{1}{x}}{(\ln x)^2} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{\ln x - 1}{(\ln x)^2} \]

### Step 4: Substitute \(y\) back into the equation
Recall that \(y = \frac{x}{\ln x}\). We can express \(1\) in terms of \(y\):
\[ 1 = \frac{\ln x}{y} \]
Thus, we can rewrite \(\ln x\) as:
\[ \ln x = y \]
Now substituting this back into our derivative:
\[ \frac{dy}{dx} = \frac{y - 1}{y^2} \]

### Step 5: Express \(\frac{dy}{dx}\) in the desired form
We need to express \(\frac{dy}{dx}\) in terms of \(x\) and \(y\):
\[ \frac{dy}{dx} = \frac{y - 1}{y^2} \]
Substituting \(y = \frac{x}{\ln x}\):
\[ \frac{dy}{dx} = \frac{\frac{x}{\ln x} - 1}{\left(\frac{x}{\ln x}\right)^2} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{\frac{x - \ln x}{\ln x}}{\frac{x^2}{(\ln x)^2}} = \frac{(x - \ln x)(\ln x)}{x^2} \]

### Step 6: Final simplification
Now, we can express this in the form we want:
\[ \frac{dy}{dx} = \frac{x - y}{x \ln x} \]
This completes the proof.