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

Explanation

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

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

### Step 2: Apply logarithmic properties
Using the property of logarithms that \(\log(a^b) = b \log(a)\) and \(\log(e^a) = a\), we can rewrite the equation:
\[ x \log y = y - x \]

### Step 3: Rearrange the equation
Rearranging gives us:
\[ x \log y + x = y \]
or
\[ y = x(1 + \log y) \]

### Step 4: Differentiate both sides with respect to \(x\)
Now, we differentiate both sides with respect to \(x\):
\[ \frac{dy}{dx} = 1 + \log y + x \cdot \frac{1}{y} \cdot \frac{dy}{dx} \]
Here, we have used the product rule on the right side.

### Step 5: Isolate \(\frac{dy}{dx}\)
Rearranging the equation gives:
\[ \frac{dy}{dx} - \frac{x}{y} \frac{dy}{dx} = 1 + \log y \]
Factoring out \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} \left(1 - \frac{x}{y}\right) = 1 + \log y \]

### Step 6: Solve for \(\frac{dy}{dx}\)
Now, solving for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{1 + \log y}{1 - \frac{x}{y}} \]
To simplify the denominator, we can rewrite it as:
\[ 1 - \frac{x}{y} = \frac{y - x}{y} \]
Thus, we have:
\[ \frac{dy}{dx} = \frac{(1 + \log y) y}{y - x} \]

### Step 7: Substitute back to find the final form
We know from our earlier rearrangement that \(y - x = x \log y\), so substituting this in gives:
\[ \frac{dy}{dx} = \frac{(1 + \log y) y}{x \log y} \]
Now, we can express \(y\) in terms of \(x\) and \(\log y\). Since \(y = x(1 + \log y)\), we can substitute \(y\) back into the equation:
\[ \frac{dy}{dx} = \frac{(1 + \log y)^2}{\log y} \]

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