Question

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

Explanation:

Step1: Differentiate the given equation

Differentiate $e^{y}-e^{x}=x - x^{3}$ with respect to $x$. Using the chain - rule, the derivative of $e^{y}$ with respect to $x$ is $e^{y}\frac{dy}{dx}$, and the derivative of $e^{x}$ is $e^{x}$, the derivative of $x$ is $1$ and the derivative of $x^{3}$ is $3x^{2}$. So we have $e^{y}\frac{dy}{dx}-e^{x}=1 - 3x^{2}$.

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

Rearrange the equation $e^{y}\frac{dy}{dx}-e^{x}=1 - 3x^{2}$ to isolate $\frac{dy}{dx}$. Add $e^{x}$ to both sides: $e^{y}\frac{dy}{dx}=1 - 3x^{2}+e^{x}$. Then $\frac{dy}{dx}=\frac{1 - 3x^{2}+e^{x}}{e^{y}}$.

Step3: Substitute the point $(0,1)$

Substitute $x = 0$ and $y = 1$ into $\frac{dy}{dx}$. When $x = 0$ and $y = 1$, we have $e^{y}=e^{1}=e$, $1-3x^{2}+e^{x}=1-3\times0^{2}+e^{0}=1 + 1=2$. So $\frac{dy}{dx}=\frac{2}{e}$.

Answer:

B. $\frac{2}{e}$