If `y=x+e ^(x),` then `((d^(2)y)/(dy ^(2)))_(x = ln 2)` is :

Options

A. `-1/9`
B. `-2/27`
C. `2/27`
D. `1/9`

Answer

To solve the problem, we need to find the second derivative of \( x \) with respect to \( y \) at \( x = \ln 2 \) given the equation \( y = x + e^x \).

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ y = x + e^x \]

2. **Differentiate both sides with respect to \( y \):**
\[ \frac{dy}{dy} = \frac{dx}{dy} + e^x \frac{dx}{dy} \]
Since \(\frac{dy}{dy} = 1\), we can rewrite the equation as:
\[ 1 = \frac{dx}{dy} (1 + e^x) \]

3. **Rearranging the equation to solve for \(\frac{dx}{dy}\):**
\[ \frac{dx}{dy} = \frac{1}{1 + e^x} \]

4. **Evaluate \(\frac{dx}{dy}\) at \( x = \ln 2 \):**
\[ \frac{dx}{dy} \bigg|_{x = \ln 2} = \frac{1}{1 + e^{\ln 2}} = \frac{1}{1 + 2} = \frac{1}{3} \]

5. **Differentiate the equation \( 1 = \frac{dx}{dy} (1 + e^x) \) again with respect to \( y \):**
\[ 0 = \frac{d^2x}{dy^2} (1 + e^x) + \frac{dx}{dy} e^x \frac{dx}{dy} \]

6. **Substituting \(\frac{dx}{dy}\) into the equation:**
\[ 0 = \frac{d^2x}{dy^2} (1 + e^x) + \left(\frac{1}{3}\right) e^x \left(\frac{1}{3}\right) \]
\[ 0 = \frac{d^2x}{dy^2} (1 + e^x) + \frac{1}{9} e^x \]

7. **Rearranging to solve for \(\frac{d^2x}{dy^2}\):**
\[ \frac{d^2x}{dy^2} (1 + e^x) = -\frac{1}{9} e^x \]
\[ \frac{d^2x}{dy^2} = -\frac{1}{9} \cdot \frac{e^x}{1 + e^x} \]

8. **Evaluate \(\frac{d^2x}{dy^2}\) at \( x = \ln 2 \):**
\[ \frac{d^2x}{dy^2} \bigg|_{x = \ln 2} = -\frac{1}{9} \cdot \frac{e^{\ln 2}}{1 + e^{\ln 2}} = -\frac{1}{9} \cdot \frac{2}{1 + 2} = -\frac{1}{9} \cdot \frac{2}{3} = -\frac{2}{27} \]

### Final Answer:
\[ \frac{d^2x}{dy^2} \bigg|_{x = \ln 2} = -\frac{2}{27} \]