Question:
If $e^{y}+x^{2}y+xy^{2}=e^{1}$, then $\frac{dy}{dx}$ at (0,1) is equal to ________.
If $e^{y}+x^{2}y+xy^{2}=e^{1}$, then $\frac{dy}{dx}$ at (0,1) is equal to ________.
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Implicitly differentiate and then plug in the point immediately to simplify algebra.
Updated On: Jun 26, 2026
- $\frac{1}{e}$
- $e$
- $-e$
- $\frac{2}{e}$
- $\frac{-1}{e}$
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The Correct Option is
Solution and Explanation
Step 1: Concept
Use implicit differentiation with respect to $x$.
Step 2: Meaning
$\frac{d}{dx}(e^y + x^2y + xy^2) = 0 \implies e^y \frac{dy}{dx} + (2xy + x^2\frac{dy}{dx}) + (y^2 + 2xy\frac{dy}{dx}) = 0$.
Step 3: Analysis
Substitute $(x,y) = (0,1)$: $e^1 \frac{dy}{dx} + (0 + 0) + (1 + 0) = 0$.
Step 4: Conclusion
$e \frac{dy}{dx} = -1 \implies \frac{dy}{dx} = -\frac{1}{e}$. Final Answer: (E)
Use implicit differentiation with respect to $x$.
Step 2: Meaning
$\frac{d}{dx}(e^y + x^2y + xy^2) = 0 \implies e^y \frac{dy}{dx} + (2xy + x^2\frac{dy}{dx}) + (y^2 + 2xy\frac{dy}{dx}) = 0$.
Step 3: Analysis
Substitute $(x,y) = (0,1)$: $e^1 \frac{dy}{dx} + (0 + 0) + (1 + 0) = 0$.
Step 4: Conclusion
$e \frac{dy}{dx} = -1 \implies \frac{dy}{dx} = -\frac{1}{e}$. Final Answer: (E)
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