Question

if (e^{2y}-e^{(y^{2}-y)}=x^{4}-x^{3}), then the value of (\frac{dy}{dx}) at the point ((1,0)) is

Explanation:

Step1: Differentiate the given equation implicitly

Given \(e^{2y}-e^{(y^{2}-y)}=x^{4}-x^{2}\). Differentiate both sides with respect to \(x\) using the chain - rule.
The derivative of the left - hand side: \(\frac{d}{dx}(e^{2y}-e^{(y^{2}-y)}) = 2e^{2y}\frac{dy}{dx}-(2y - 1)e^{(y^{2}-y)}\frac{dy}{dx}\).
The derivative of the right - hand side: \(\frac{d}{dx}(x^{4}-x^{2})=4x^{3}-2x\).
So, \((2e^{2y}-(2y - 1)e^{(y^{2}-y)})\frac{dy}{dx}=4x^{3}-2x\).
Then \(\frac{dy}{dx}=\frac{4x^{3}-2x}{2e^{2y}-(2y - 1)e^{(y^{2}-y)}}\).

Step2: Substitute \(x = 1\) and \(y = 0\)

Substitute \(x = 1\) and \(y = 0\) into \(\frac{dy}{dx}\).
When \(x = 1\) and \(y = 0\), the numerator \(4x^{3}-2x=4\times1^{3}-2\times1=4 - 2=2\).
The denominator \(2e^{2y}-(2y - 1)e^{(y^{2}-y)}=2e^{2\times0}-(2\times0 - 1)e^{(0^{2}-0)}=2\times1-(- 1)\times1=2 + 1=3\).
So \(\frac{dy}{dx}=\frac{2}{3}\).

Answer:

C. \(\frac{2}{3}\)