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

Explanation

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

### Step 1: Differentiate both sides
We start with the equation:
\[ x = e^{\frac{x}{y}} \]
Differentiating both sides with respect to \(x\):
\[ \frac{d}{dx}(x) = \frac{d}{dx}\left(e^{\frac{x}{y}}\right) \]
This gives us:
\[ 1 = \frac{d}{dx}\left(e^{\frac{x}{y}}\right) \]

### Step 2: Apply the chain rule
Using the chain rule on the right side:
\[ 1 = e^{\frac{x}{y}} \cdot \frac{d}{dx}\left(\frac{x}{y}\right) \]

### Step 3: Differentiate \(\frac{x}{y}\) using the quotient rule
Now we need to differentiate \(\frac{x}{y}\) using the quotient rule:
\[ \frac{d}{dx}\left(\frac{x}{y}\right) = \frac{y \cdot \frac{d}{dx}(x) - x \cdot \frac{dy}{dx}}{y^2} \]
Substituting \(\frac{d}{dx}(x) = 1\):
\[ \frac{d}{dx}\left(\frac{x}{y}\right) = \frac{y \cdot 1 - x \cdot \frac{dy}{dx}}{y^2} = \frac{y - x \frac{dy}{dx}}{y^2} \]

### Step 4: Substitute back into the equation
Substituting this back into our equation:
\[ 1 = e^{\frac{x}{y}} \cdot \frac{y - x \frac{dy}{dx}}{y^2} \]

### Step 5: Rearranging the equation
Now, we can rearrange this equation:
\[ y^2 = e^{\frac{x}{y}} (y - x \frac{dy}{dx}) \]
Expanding this gives:
\[ y^2 = y e^{\frac{x}{y}} - x e^{\frac{x}{y}} \frac{dy}{dx} \]

### Step 6: Isolate \(\frac{dy}{dx}\)
Now, we isolate \(\frac{dy}{dx}\):
\[ x e^{\frac{x}{y}} \frac{dy}{dx} = y e^{\frac{x}{y}} - y^2 \]
\[ \frac{dy}{dx} = \frac{y e^{\frac{x}{y}} - y^2}{x e^{\frac{x}{y}}} \]

### Step 7: Simplify the expression
Factoring out \(y\) from the numerator:
\[ \frac{dy}{dx} = \frac{y(e^{\frac{x}{y}} - y)}{x e^{\frac{x}{y}}} \]

### Step 8: Substitute \(e^{\frac{x}{y}} = x\)
From our original equation \(x = e^{\frac{x}{y}}\), we can substitute \(e^{\frac{x}{y}}\) with \(x\):
\[ \frac{dy}{dx} = \frac{y(x - y)}{x \cdot x} = \frac{y(x - y)}{x^2} \]

### Step 9: Rewrite using \(\log x\)
Since \(y = x \log x\) (from \(x = e^{\frac{x}{y}}\)), we can express \(y\) in terms of \(x\):
\[ \frac{dy}{dx} = \frac{x - y}{x \log x} \]

Thus, we have proved that:
\[ \frac{dy}{dx} = \frac{x - y}{x \log x} \]