Questions: Consider the curve defined by (x^2=e^x-y) for (x>0). At what value of (x) does the curve have a horizontal tangent? (A) 1 (B) (sqrt2) C 2 D There is no such value of (x).

To find the value of \( x \) where the curve \( x^2 = e^{x-y} \) has a horizontal tangent, we need to determine where the derivative of \( y \) with respect to \( x \) is zero. This involves implicit differentiation of the given equation and solving for \( \frac{dy}{dx} = 0 \).

Given the curve \( x^2 = e^{x-y} \), we differentiate both sides with respect to \( x \) implicitly: \[ \frac{d}{dx}(x^2) = \frac{d}{dx}(e^{x-y}) \] \[ 2x = e^{x-y} \left(1 - \frac{dy}{dx}\right) \]

Rearrange the equation to solve for \(\frac{dy}{dx}\): \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} \left(1 - \frac{dy}{dx}\right) \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - e^{x-y} \frac{dy}{dx} \] \[ 2x = e^{x-y} - 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