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The curve y-e^(xy)+x=0 has a vertical ta...

The curve `y-e^(xy)+x=0` has a vertical tangent at the point :

A

(1, 1)

B

at no point

C

(0, 1)

D

(1, 0)

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To find the point where the curve \( y - e^{xy} + x = 0 \) has a vertical tangent, we need to determine where the derivative \( \frac{dy}{dx} \) is undefined, which typically occurs when the denominator of the derivative equals zero. ### Step-by-Step Solution: 1. **Differentiate the given equation**: We start with the equation of the curve: \[ y - e^{xy} + x = 0 \] We will differentiate both sides with respect to \( x \). 2. **Apply implicit differentiation**: Differentiating \( y \) gives \( \frac{dy}{dx} \). For \( e^{xy} \), we use the chain rule: \[ \frac{d}{dx}(e^{xy}) = e^{xy} \cdot \left( y + x \frac{dy}{dx} \right) \] The derivative of \( x \) is simply \( 1 \). Thus, differentiating the entire equation gives: \[ \frac{dy}{dx} - e^{xy} \left( y + x \frac{dy}{dx} \right) + 1 = 0 \] 3. **Rearranging the equation**: Rearranging the equation to isolate \( \frac{dy}{dx} \): \[ \frac{dy}{dx} - e^{xy} y - x e^{xy} \frac{dy}{dx} + 1 = 0 \] Combine the \( \frac{dy}{dx} \) terms: \[ \frac{dy}{dx} (1 - x e^{xy}) = e^{xy} y - 1 \] 4. **Solve for \( \frac{dy}{dx} \)**: Now, we can express \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{e^{xy} y - 1}{1 - x e^{xy}} \] 5. **Determine where the tangent is vertical**: A vertical tangent occurs when the denominator is zero: \[ 1 - x e^{xy} = 0 \implies x e^{xy} = 1 \] 6. **Substituting points**: We need to find the points \( (x, y) \) that satisfy this equation. We will check the provided options to see which one satisfies this condition. - For point A: \( (1, 1) \) \[ 1 \cdot e^{1 \cdot 1} = e \quad (\text{not } 1) \] - For point B: \( (0, 0) \) \[ 0 \cdot e^{0 \cdot 0} = 0 \quad (\text{not } 1) \] - For point C: \( (1, 0) \) \[ 1 \cdot e^{1 \cdot 0} = 1 \quad (\text{satisfies}) \] - For point D: \( (1, 0) \) \[ 1 \cdot e^{1 \cdot 0} = 1 \quad (\text{satisfies}) \] 7. **Conclusion**: The point where the curve has a vertical tangent is: \[ (1, 0) \]

To find the point where the curve \( y - e^{xy} + x = 0 \) has a vertical tangent, we need to determine where the derivative \( \frac{dy}{dx} \) is undefined, which typically occurs when the denominator of the derivative equals zero. ### Step-by-Step Solution: 1. **Differentiate the given equation**: We start with the equation of the curve: \[ y - e^{xy} + x = 0 ...
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