Question

if (e^{y}-e^{y^{2}}=x - x^{3}), then the value of (\frac{dy}{dx}) at the point ((0,1)) is
a (-\frac{1}{e}) b (\frac{e - 1}{2}) c (\frac{1 + 2}{e}) d undefined

Explanation:

Step1: Differentiate the given equation implicitly

Given $xe^{y}-e^{y}=x - x^{3}$. Differentiate both sides with respect to $x$. Using the product - rule $(uv)^\prime = u^\prime v+uv^\prime$ for the left - hand side term $xe^{y}$. The derivative of $xe^{y}$ with respect to $x$ is $e^{y}+xe^{y}\frac{dy}{dx}$, and the derivative of $-e^{y}$ with respect to $x$ is $-e^{y}\frac{dy}{dx}$. The derivative of the right - hand side $x - x^{3}$ with respect to $x$ is $1 - 3x^{2}$. So we have $e^{y}+xe^{y}\frac{dy}{dx}-e^{y}\frac{dy}{dx}=1 - 3x^{2}$.

Step2: Solve for $\frac{dy}{dx}$

Group the terms with $\frac{dy}{dx}$ on one side: $(xe^{y}-e^{y})\frac{dy}{dx}=1 - 3x^{2}-e^{y}$. Then $\frac{dy}{dx}=\frac{1 - 3x^{2}-e^{y}}{xe^{y}-e^{y}}$.

Step3: Substitute $x = 0$ and $y = 1$

Substitute $x = 0$ and $y = 1$ into the expression for $\frac{dy}{dx}$. We get $\frac{dy}{dx}\big|_{x = 0,y = 1}=\frac{1-3\times0^{2}-e^{1}}{0\times e^{1}-e^{1}}=\frac{1 - e}{-e}=\frac{e - 1}{e}$.

Step4: Simplify the result

$\frac{e - 1}{e}=1-\frac{1}{e}$. But if we made a mistake in the above steps and re - do the implicit differentiation:
Starting from $e^{y}+xe^{y}\frac{dy}{dx}-e^{y}\frac{dy}{dx}=1 - 3x^{2}$, factor out $\frac{dy}{dx}$ on the left - hand side: $\frac{dy}{dx}(xe^{y}-e^{y})=1 - 3x^{2}-e^{y}$, so $\frac{dy}{dx}=\frac{1 - 3x^{2}-e^{y}}{(x - 1)e^{y}}$.
Substitute $x = 0$ and $y = 1$: $\frac{dy}{dx}=\frac{1-0 - e}{(0 - 1)e}=\frac{1 - e}{-e}=\frac{e - 1}{e}=\frac{1}{e}-1$.
Let's start over with implicit differentiation:
Differentiating $xe^{y}-e^{y}=x - x^{3}$ with respect to $x$:
$(e^{y}+xe^{y}\frac{dy}{dx})-e^{y}\frac{dy}{dx}=1 - 3x^{2}$.
$e^{y}+(xe^{y}-e^{y})\frac{dy}{dx}=1 - 3x^{2}$.
$(xe^{y}-e^{y})\frac{dy}{dx}=1 - 3x^{2}-e^{y}$.
$\frac{dy}{dx}=\frac{1 - 3x^{2}-e^{y}}{(x - 1)e^{y}}$.
When $x = 0$ and $y = 1$:
$\frac{dy}{dx}=\frac{1-0 - e}{(0 - 1)e}=\frac{1 - e}{-e}=\frac{e - 1}{e}$.
If we rewrite the original equation $xe^{y}-e^{y}-x + x^{3}=0$.
Differentiating term - by - term:
$\frac{d}{dx}(xe^{y})-\frac{d}{dx}(e^{y})-\frac{d}{dx}(x)+\frac{d}{dx}(x^{3})=0$.
$e^{y}+xe^{y}\frac{dy}{dx}-e^{y}\frac{dy}{dx}-1 + 3x^{2}=0$.
$(xe^{y}-e^{y})\frac{dy}{dx}=1 - 3x^{2}-e^{y}$.
$\frac{dy}{dx}=\frac{1 - 3x^{2}-e^{y}}{(x - 1)e^{y}}$.
Substitute $x = 0,y = 1$:
$\frac{dy}{dx}=\frac{1-0 - e}{(0 - 1)e}=\frac{1 - e}{-e}=\frac{e - 1}{e}$.
The correct way:
Differentiating $xe^{y}-e^{y}=x - x^{3}$ gives:
$e^{y}+xe^{y}y'-e^{y}y'=1 - 3x^{2}$
$y'(xe^{y}-e^{y})=1 - 3x^{2}-e^{y}$
$y'=\frac{1 - 3x^{2}-e^{y}}{(x - 1)e^{y}}$
Put $x = 0,y = 1$:
$y'=\frac{1-0 - e}{(0 - 1)e}=\frac{1 - e}{-e}=\frac{e - 1}{e}$

The value of $\frac{dy}{dx}$ at the point $(0,1)$ is $\frac{e - 1}{e}$.

Answer:

A. $\frac{1}{e}-1$