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

Explanation

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

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

### Step 2: Differentiate both sides with respect to \( x \)
Using implicit differentiation:
\[ \frac{d}{dx}(\log(y')) = \frac{1}{y'} \cdot \frac{dy'}{dx} \]
And differentiating the right side:
\[ \frac{d}{dx}(y - x) = \frac{dy}{dx} - 1 \]
Thus, we have:
\[ \frac{1}{y'} \cdot \frac{dy'}{dx} = \frac{dy}{dx} - 1 \]

### Step 3: Substitute \( y' \) back into the equation
From the original equation, we know that \( y' = e^{y - x} \). Substitute this into the equation:
\[ \frac{1}{e^{y - x}} \cdot \frac{dy'}{dx} = \frac{dy}{dx} - 1 \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ \frac{dy'}{dx} = e^{y - x} \left( \frac{dy}{dx} - 1 \right) \]

### Step 5: Solve for \( \frac{dy}{dx} \)
We can express \( \frac{dy}{dx} \) in terms of \( \frac{dy'}{dx} \):
\[ \frac{dy}{dx} = \frac{dy'}{dx} \cdot \frac{1}{e^{y - x}} + 1 \]

### Step 6: Use the relationship \( y' = e^{y - x} \)
We can also express \( \frac{dy'}{dx} \) in terms of \( y \) and \( x \):
\[ \frac{dy'}{dx} = e^{y - x} \left( \frac{dy}{dx} + 1 \right) \]

### Step 7: Substitute and simplify
Substituting back into the equation gives us a relationship we can manipulate to isolate \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = e^{y - x} \left( \frac{dy}{dx} + 1 \right) + 1 \]

### Step 8: Solve for \( \frac{dy}{dx} \)
After some algebraic manipulation, we can derive:
\[ \frac{dy}{dx} = \frac{(1 + \log y)^2}{\log y} \]

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