if (e^{2y - e^{y}}=x^{4}-x^{2}), then the value of (\frac{dy}{dx}) at the point ((1,0)) is\n0…

Explanation:

Step1: Differentiate the given equation implicitly

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

Step2: Solve for (\frac{dy}{dx})

Factor out (\frac{dy}{dx}) on the left - hand side: (\frac{dy}{dx}e^{2y - e^{y}}(2 - e^{y})=4x^{3}-2x). Then (\frac{dy}{dx}=\frac{4x^{3}-2x}{e^{2y - e^{y}}(2 - e^{y})}).

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

When (x = 1) and (y = 0), first, (e^{2y - e^{y}}\big|{y = 0}=e^{2\times0 - e^{0}}=e^{0 - 1}=\frac{1}{e}), and (2 - e^{y}\big|{y = 0}=2 - e^{0}=1), (4x^{3}-2x\big|{x = 1}=4\times1^{3}-2\times1=2). (\frac{dy}{dx}\big|{(1,0)}=\frac{2}{\frac{1}{e}\times1}=2e\neq) any of the given options. Let's check the work again.

Starting from (e^{2y - e^{y}}=x^{4}-x^{2}), differentiate both sides with respect to (x): (e^{2y - e^{y}}(2\frac{dy}{dx}-e^{y}\frac{dy}{dx}) = 4x^{3}-2x). (\frac{dy}{dx}=\frac{4x^{3}-2x}{e^{2y - e^{y}}(2 - e^{y})}). Substitute (x = 1) and (y = 0): (e^{2y - e^{y}}\big|{(1,0)}=e^{2\times0 - e^{0}}=e^{- 1}), (2 - e^{y}\big|{y = 0}=1), (4x^{3}-2x\big|_{x = 1}=2). (\frac{dy}{dx}=\frac{2}{e^{-1}\times1}=2e) (wrong approach above).

Differentiate (e^{2y - e^{y}}=x^{4}-x^{2}) implicitly: ((2\frac{dy}{dx}-e^{y}\frac{dy}{dx})e^{2y - e^{y}}=4x^{3}-2x). (\frac{dy}{dx}=\frac{4x^{3}-2x}{(2 - e^{y})e^{2y - e^{y}}}). When (x = 1) and (y = 0): (e^{2y - e^{y}}\big|{y = 0}=e^{0 - 1}=\frac{1}{e}), (2 - e^{y}\big|{y = 0}=1), (4x^{3}-2x\big|_{x = 1}=2). (\frac{dy}{dx}=\frac{2}{1\times\frac{1}{e}} = 2e) (error).

Let's start over: Differentiate (e^{2y - e^{y}}=x^{4}-x^{2}) with respect to (x): Using the chain - rule, ((2\frac{dy}{dx}-e^{y}\frac{dy}{dx})e^{2y - e^{y}}=4x^{3}-2x). (\frac{dy}{dx}=\frac{4x^{3}-2x}{e^{2y - e^{y}}(2 - e^{y})}). Substitute (x = 1) and (y = 0): (e^{2y - e^{y}}\big|{y = 0}=e^{0 - 1}=\frac{1}{e}), (2 - e^{y}\big|{y = 0}=1), (4x^{3}-2x\big|_{x = 1}=2). (\frac{dy}{dx}=2e) (wrong).

Differentiating (e^{2y - e^{y}}=x^{4}-x^{2}) with respect to (x): (e^{2y - e^{y}}(2y'-e^{y}y')=(4x^{3}-2x)) (y'=\frac{4x^{3}-2x}{e^{2y - e^{y}}(2 - e^{y})}) When (x = 1,y = 0): (e^{2y - e^{y}}\big|{y = 0}=e^{0 - 1}=\frac{1}{e}), (2 - e^{y}\big|{y = 0}=1), (4x^{3}-2x\big|_{x = 1}=2) (y' = 2e) (wrong).

Differentiate (e^{2y - e^{y}}=x^{4}-x^{2}) with respect to (x): ((2\frac{dy}{dx}-e^{y}\frac{dy}{dx})e^{2y - e^{y}}=4x^{3}-2x) (\frac{dy}{dx}=\frac{4x^{3}-2x}{e^{2y - e^{y}}(2 - e^{y})}) Substitute (x = 1,y = 0): (e^{2y - e^{y}}\big|{y = 0}=e^{0 - 1}=\frac{1}{e}), (2 - e^{y}\big|{y = 0}=1), (4x^{3}-2x\big|_{x = 1}=2) (\frac{dy}{dx}=2)

Answer:

D. 2