If `e^y+xy=e` then the value of `(d^2y)/(dx^2)` for `x=0` is

Options

A. `1//e`
B. `1//e^(2)`
C. `1//e^(3)`
D. e

Answer

To solve the problem, we start with the equation given:

\[ e^y + xy = e \]

### Step 1: Differentiate both sides with respect to \( x \)

Differentiating both sides gives us:

\[ \frac{d}{dx}(e^y) + \frac{d}{dx}(xy) = \frac{d}{dx}(e) \]

Since \( e \) is a constant, its derivative is 0. Now we apply the chain rule and the product rule:

1. For \( \frac{d}{dx}(e^y) \), we use the chain rule:
\[ \frac{d}{dx}(e^y) = e^y \frac{dy}{dx} \]

2. For \( \frac{d}{dx}(xy) \), we use the product rule:
\[ \frac{d}{dx}(xy) = x \frac{dy}{dx} + y \]

Putting this all together, we have:

\[ e^y \frac{dy}{dx} + x \frac{dy}{dx} + y = 0 \]

### Step 2: Rearranging the equation

We can factor out \( \frac{dy}{dx} \):

\[ (e^y + x) \frac{dy}{dx} + y = 0 \]

Now, solving for \( \frac{dy}{dx} \):

\[ (e^y + x) \frac{dy}{dx} = -y \]

Thus,

\[ \frac{dy}{dx} = \frac{-y}{e^y + x} \]

### Step 3: Evaluate at \( x = 0 \)

Substituting \( x = 0 \) into the original equation to find \( y \):

\[ e^y + 0 \cdot y = e \]
\[ e^y = e \]
\[ y = 1 \]

Now substituting \( x = 0 \) and \( y = 1 \) into the derivative:

\[ \frac{dy}{dx} = \frac{-1}{e^1 + 0} = \frac{-1}{e} \]

### Step 4: Differentiate again to find \( \frac{d^2y}{dx^2} \)

Now we differentiate \( \frac{dy}{dx} \) again with respect to \( x \):

Using the quotient rule:

\[ \frac{d^2y}{dx^2} = \frac{(e^y + x) \frac{d}{dx}(-y) - (-y) \frac{d}{dx}(e^y + x)}{(e^y + x)^2} \]

Calculating \( \frac{d}{dx}(-y) = -\frac{dy}{dx} \) and \( \frac{d}{dx}(e^y + x) = e^y \frac{dy}{dx} + 1 \):

Substituting these into the equation gives:

\[ \frac{d^2y}{dx^2} = \frac{(e^y + x)(-\frac{dy}{dx}) - (-y)(e^y \frac{dy}{dx} + 1)}{(e^y + x)^2} \]

### Step 5: Substitute \( x = 0 \) and \( y = 1 \)

Substituting \( x = 0 \) and \( y = 1 \):

\[ \frac{d^2y}{dx^2} = \frac{(e^1 + 0)(\frac{1}{e}) - (-1)(e^1 \cdot \frac{-1}{e} + 1)}{(e + 0)^2} \]

This simplifies to:

\[ \frac{d^2y}{dx^2} = \frac{(e)(\frac{-1}{e}) - (-1)(-1 + 1)}{e^2} \]

Calculating this gives:

\[ \frac{d^2y}{dx^2} = \frac{-1}{e} \]

Thus, the value of \( \frac{d^2y}{dx^2} \) at \( x = 0 \) is:

\[ \frac{d^2y}{dx^2} = -\frac{1}{e} \]