Question:
If $y = e^{-x^2}$, then $\dfrac{d^2y}{dx^2} + 2x\dfrac{dy}{dx}$ is equal to:
If $y = e^{-x^2}$, then $\dfrac{d^2y}{dx^2} + 2x\dfrac{dy}{dx}$ is equal to:
Show Hint
Express derivatives in terms of $y$ to simplify expressions quickly.
Updated On: Apr 24, 2026
- $2y$
- $-2y$
- $-\frac{y}{2}$
- $-y$
- $y$
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The Correct Option is B
Solution and Explanation
Concept:
• Use chain rule for exponential functions
Step 1: First derivative
\[ \frac{dy}{dx} = e^{-x^2}(-2x) = -2xy \]
Step 2: Second derivative
\[ \frac{d^2y}{dx^2} = -2y -2x\frac{dy}{dx} \]
Step 3: Substitute
\[ \frac{d^2y}{dx^2} + 2x\frac{dy}{dx} = -2y \] Final Conclusion:
\[ = -2y \]
• Use chain rule for exponential functions
Step 1: First derivative
\[ \frac{dy}{dx} = e^{-x^2}(-2x) = -2xy \]
Step 2: Second derivative
\[ \frac{d^2y}{dx^2} = -2y -2x\frac{dy}{dx} \]
Step 3: Substitute
\[ \frac{d^2y}{dx^2} + 2x\frac{dy}{dx} = -2y \] Final Conclusion:
\[ = -2y \]
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