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})