If ` y^(x)=e^(y-x)`, then prove that `(dy)/(dx) = ((1+logy)^(2))/(logy)`

Explanation

To prove that \(\frac{dy}{dx} = \frac{(1 + \log y)^2}{\log y}\) given the equation \(y^x = e^{y - x}\), we will follow these steps:

### Step 1: Take the natural logarithm of both sides
Starting with the equation:
\[ y^x = e^{y - x} \]
Taking the natural logarithm of both sides:
\[ \log(y^x) = \log(e^{y - x}) \]

### Step 2: Simplify using logarithmic properties
Using the properties of logarithms, we can simplify both sides:
\[ x \log y = y - x \]

### Step 3: Differentiate both sides with respect to \(x\)
Now, we differentiate both sides with respect to \(x\):
\[ \frac{d}{dx}(x \log y) = \frac{d}{dx}(y - x) \]
Using the product rule on the left side:
\[ \log y + x \frac{1}{y} \frac{dy}{dx} = \frac{dy}{dx} - 1 \]

### Step 4: Rearrange the equation
Rearranging the equation gives:
\[ x \frac{1}{y} \frac{dy}{dx} - \frac{dy}{dx} = -1 - \log y \]
Factoring out \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} \left( x \frac{1}{y} - 1 \right) = -1 - \log y \]

### Step 5: Solve for \(\frac{dy}{dx}\)
Now, we can solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{-1 - \log y}{x \frac{1}{y} - 1} \]

### Step 6: Simplify the expression
To simplify this expression, we can rewrite it as:
\[ \frac{dy}{dx} = \frac{-(1 + \log y)}{\frac{x}{y} - 1} \]
Multiplying numerator and denominator by \(-1\):
\[ \frac{dy}{dx} = \frac{1 + \log y}{1 - \frac{x}{y}} = \frac{(1 + \log y)}{\frac{y - x}{y}} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{y(1 + \log y)}{y - x} \]

### Step 7: Use the relation \(y - x = x \log y\)
From our earlier work, we know that:
\[ y - x = x \log y \]
Substituting this into our expression for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{y(1 + \log y)}{x \log y} \]

### Step 8: Rearranging the expression
Now, we can rearrange it to match the required form:
\[ \frac{dy}{dx} = \frac{y(1 + \log y)}{x \log y} = \frac{(1 + \log y)^2}{\log y} \]

### Conclusion
Thus, we have proved that:
\[ \frac{dy}{dx} = \frac{(1 + \log y)^2}{\log y} \]