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 - e}{2e})
c (\frac{1 + 2e}{e^{2}})
d undefined

Explanation:

Step1: Differentiate the given function

Given \(y = e^{y}-x - x^{3}\). Differentiate both sides with respect to \(x\) using implicit - differentiation. The derivative of \(y\) with respect to \(x\) is \(\frac{dy}{dx}\), the derivative of \(e^{y}\) with respect to \(x\) is \(e^{y}\frac{dy}{dx}\) (by the chain - rule), the derivative of \(x\) with respect to \(x\) is \(1\), and the derivative of \(x^{3}\) with respect to \(x\) is \(3x^{2}\). So we have \(\frac{dy}{dx}=e^{y}\frac{dy}{dx}-1 - 3x^{2}\).

Step2: Isolate \(\frac{dy}{dx}\) terms

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

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

Substitute \(x = 0\) and \(y = 1\) into \(\frac{dy}{dx}\). We get \(\frac{dy}{dx}\big|_{(0,1)}=\frac{-1-3\times0^{2}}{1 - e^{1}}=\frac{-1}{1 - e}=\frac{1}{e - 1}\).

Answer:

B. \(\frac{1}{e - 1}\)