Question

3 mark for review if (e^{y}-e^{y}x^{3}=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 implicitly

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

Step2: Factor out $\frac{dy}{dx}$

Factor out $\frac{dy}{dx}$ on the left - hand side: $\frac{dy}{dx}(e^{y}-1)=1 - 3x^{2}$. Then $\frac{dy}{dx}=\frac{1 - 3x^{2}}{e^{y}-1}$.

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

Substitute $x = 0$ and $y = 1$ into the derivative formula. When $x = 0$ and $y = 1$, we get $\frac{dy}{dx}=\frac{1-3\times0^{2}}{e^{1}-1}=\frac{1}{e - 1}=-\frac{1}{1 - e}$.

Answer:

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