Question

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

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

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

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

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

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

Answer:

A. $\frac{1}{e+\frac{1}{e}}$ (equivalent to $\frac{e}{e^{2}+1}$)