Consider the equation `x^(y)=e^(x-y)`
What is `(d^(2)y)/(dx^(2))` at `x=1` equal to ?

Options

A. 0
B. 1
C. 2
D. 4

Answer

To solve the problem, we need to find the second derivative of \( y \) with respect to \( x \) from the equation \( x^y = e^{(x - y)} \) at \( x = 1 \). Here’s a step-by-step solution:

### Step 1: Take the natural logarithm of both sides
We start with the equation:
\[ x^y = e^{(x - y)} \]
Taking the natural logarithm on both sides gives:
\[ \ln(x^y) = \ln(e^{(x - y)}) \]
Using the properties of logarithms, we can simplify this to:
\[ y \ln(x) = x - y \]

### Step 2: Rearranging the equation
Rearranging the equation, we get:
\[ y \ln(x) + y = x \]
Factoring out \( y \) gives:
\[ y(\ln(x) + 1) = x \]
Thus, we can express \( y \) as:
\[ y = \frac{x}{\ln(x) + 1} \]

### Step 3: Differentiate \( y \) with respect to \( x \) to find \( \frac{dy}{dx} \)
Now we differentiate \( y \):
\[ \frac{dy}{dx} = \frac{(\ln(x) + 1) \cdot 1 - x \cdot \frac{1}{x}}{(\ln(x) + 1)^2} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{\ln(x) + 1 - 1}{(\ln(x) + 1)^2} = \frac{\ln(x)}{(\ln(x) + 1)^2} \]

### Step 4: Differentiate again to find \( \frac{d^2y}{dx^2} \)
Next, we differentiate \( \frac{dy}{dx} \):
Using the quotient rule:
\[ \frac{d^2y}{dx^2} = \frac{(\ln(x) + 1)^2 \cdot \frac{1}{x} - \ln(x) \cdot 2(\ln(x) + 1) \cdot \frac{1}{x}}{(\ln(x) + 1)^4} \]
Simplifying this gives:
\[ \frac{d^2y}{dx^2} = \frac{\frac{(\ln(x) + 1)^2}{x} - \frac{2\ln(x)(\ln(x) + 1)}{x}}{(\ln(x) + 1)^4} \]
Combining the terms in the numerator:
\[ \frac{d^2y}{dx^2} = \frac{(\ln(x) + 1)^2 - 2\ln(x)(\ln(x) + 1)}{x(\ln(x) + 1)^4} \]
This simplifies to:
\[ \frac{d^2y}{dx^2} = \frac{1 - \ln(x)}{x(\ln(x) + 1)^3} \]

### Step 5: Evaluate at \( x = 1 \)
Now we evaluate \( \frac{d^2y}{dx^2} \) at \( x = 1 \):
\[ \frac{d^2y}{dx^2} \bigg|_{x=1} = \frac{1 - \ln(1)}{1(\ln(1) + 1)^3} \]
Since \( \ln(1) = 0 \):
\[ \frac{d^2y}{dx^2} \bigg|_{x=1} = \frac{1 - 0}{1(0 + 1)^3} = \frac{1}{1} = 1 \]

### Final Answer
Thus, the value of \( \frac{d^2y}{dx^2} \) at \( x = 1 \) is:
\[ \boxed{1} \]