Question

If \(e^y + xy = e\) the ordered pair \(\left(\frac{dy}{dx},  \frac{d^2y}{dx^2}\right)\) at \(x = 0\) is equal to

• A. \(\left( \frac{1}{e}, \frac{1}{e^2} \right)\)
• B. \(\left( \frac{1}{e}, -\frac{1}{e^2} \right)\)
• C. \(\left( -\frac{1}{e}, -\frac{1}{e^2} \right)\)
• D. \(\left( -\frac{1}{e}, \frac{1}{e^2} \right)\)

Answer 1

The Correct Option is D

We are given: \[ e^y + xy = e \] 

Step 1: Differentiate implicitly with respect to \(x\)
\[ \frac{d}{dx}(e^y) + \frac{d}{dx}(xy) = \frac{d}{dx}(e) \] \[ e^y \frac{dy}{dx} + y + x \frac{dy}{dx} = 0 \] Group terms: \[ (e^y + x)\frac{dy}{dx} = -y \Rightarrow \frac{dy}{dx} = \frac{-y}{e^y + x} \] 

Step 2: Find value of \(y\) at \(x = 0\)
From original equation: \[ e^y + 0 = e \Rightarrow e^y = e \Rightarrow y = 1 \] Substitute \(x = 0, y = 1\) into \(\frac{dy}{dx}\): \[ \frac{dy}{dx} = \frac{-1}{e^1 + 0} = \frac{-1}{e} \]

 Step 3: Differentiate again to find \(\frac{d^2y}{dx^2}\)
Differentiate: \[ \frac{dy}{dx} = \frac{-y}{e^y + x} \] Use quotient rule: \[ \frac{d^2y}{dx^2} = \frac{-(e^y + x)\frac{dy}{dx} + y \cdot \left( e^y \frac{dy}{dx} + 1 \right)}{(e^y + x)^2} \] At \(x = 0, y = 1, \frac{dy}{dx} = -\frac{1}{e}\), compute numerator: \[ -(e + 0)\left(-\frac{1}{e}\right) + 1 \cdot \left( e \cdot \left(-\frac{1}{e}\right) + 1 \right) = e \cdot \frac{1}{e} + ( -1 + 1 ) = 1 + 0 = 1 \] Denominator: \[ (e^y + x)^2 = (e + 0)^2 = e^2 \] So: \[ \frac{d^2y}{dx^2} = \frac{1}{e^2} \] 

Final answer: \(\left( -\frac{1}{e}, \frac{1}{e^2} \right)\)

Answer 2

Given the equation \(e^y + xy = e\). 

First, let's find the value of \(y\) when \(x = 0\). Substitute \(x = 0\) into the equation:

\(e^y + (0)y = e\)

\(e^y = e\)

This implies \(y = 1\) when \(x = 0\).

Next, we differentiate the given equation implicitly with respect to \(x\):

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

Using the chain rule for \(e^y\) and the product rule for \(xy\):

\(\frac{d}{dx}(e^y) + \frac{d}{dx}(xy) = 0\)

\(e^y \frac{dy}{dx} + \left(1 \cdot y + x \cdot \frac{dy}{dx}\right) = 0\)

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

Rearrange the terms to solve for \(\frac{dy}{dx}\):

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

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

Now, evaluate \(\frac{dy}{dx}\) at the point \((x, y) = (0, 1)\):

\(\frac{dy}{dx}\bigg|_{(0,1)} = -\frac{1}{e^1 + 0} = \mathbf{-\frac{1}{e}}\)

Next, we find the second derivative, \(\frac{d^2y}{dx^2}\), by differentiating the expression for \(\frac{dy}{dx}\) with respect to \(x\):

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

Using the quotient rule:

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

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

Now, evaluate \(\frac{d^2y}{dx^2}\) at the point \((x, y) = (0, 1)\) and using \(\frac{dy}{dx} = -\frac{1}{e}\):

\(\frac{d^2y}{dx^2}\bigg|_{(0,1)} = - \frac{(e^1 + 0)(-\frac{1}{e}) - 1(e^1 (-\frac{1}{e}) + 1)}{(e^1 + 0)^2}\)

\(\frac{d^2y}{dx^2}\bigg|_{(0,1)} = - \frac{e(-\frac{1}{e}) - 1(e(-\frac{1}{e}) + 1)}{e^2}\)

\(\frac{d^2y}{dx^2}\bigg|_{(0,1)} = - \frac{-1 - 1(-1 + 1)}{e^2}\)

\(\frac{d^2y}{dx^2}\bigg|_{(0,1)} = - \frac{-1 - 1(0)}{e^2}\)

\(\frac{d^2y}{dx^2}\bigg|_{(0,1)} = - \frac{-1}{e^2}\)

\(\frac{d^2y}{dx^2}\bigg|_{(0,1)} = \mathbf{\frac{1}{e^2}}\)

Therefore, the ordered pair \(\left(\frac{dy}{dx}, \frac{d^2y}{dx^2}\right)\) at \(x = 0\) is \(\left(-\frac{1}{e}, \frac{1}{e^2}\right)\).

Comparing this with the given options, the correct option is:

\(\left( -\frac{1}{e}, \frac{1}{e^2} \right)\)