Question

3 mark for review 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 $\frac{e - 1}{2e}$ c $\frac{1 + 2e}{e}$ d undefined

Explanation:

Step1: Differentiate $e^{y}=x - x^{3}$ implicitly.

Differentiating both sides with respect to $x$, we use the chain - rule on the left - hand side. The derivative of $e^{y}$ with respect to $x$ is $e^{y}\frac{dy}{dx}$, and the derivative of $x - x^{3}$ with respect to $x$ is $1-3x^{2}$. So, $e^{y}\frac{dy}{dx}=1 - 3x^{2}$.

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

We get $\frac{dy}{dx}=\frac{1 - 3x^{2}}{e^{y}}$.

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

When $x = 0$ and $y = 1$, we substitute these values into the derivative formula. Since $x = 0$, then $1-3x^{2}=1-3\times0^{2}=1$, and since $y = 1$, $e^{y}=e^{1}=e$. So, $\frac{dy}{dx}=\frac{1}{e}$.

Answer:

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