If $y = y(x)$ is an implicit function of $x$ such that $\log_e (x + y) = 4xy$, then $\frac{d^2y}{dx^2}$ at $x = 0$ is equal to _________.
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When finding higher derivatives of implicit functions at a specific point, always solve for $y, y', y'', \dots$ sequentially by substituting the numerical values at each step.
To solve the problem of finding \(\frac{d^2y}{dx^2}\) when \(x=0\) for the given implicit function \(\log_e(x+y)=4xy\), follow these steps: 1. **Implicit Differentiation**: Differentiate the equation \(\log_e(x+y)=4xy\) with respect to \(x\) to find \(\frac{dy}{dx}\). 2. The derivative of the left-hand side using the chain rule is \(\frac{1}{x+y}(1+\frac{dy}{dx})\). 3. The derivative of the right-hand side is \(4x\frac{dy}{dx} + 4y\). 4. Equate the derivatives: \(\frac{1}{x+y}(1+\frac{dy}{dx}) = 4x\frac{dy}{dx} + 4y\). 5. **Solving for \(\frac{dy}{dx}\)**: Rearrange to get: \((1+\frac{dy}{dx}) = (x+y)(4x\frac{dy}{dx} + 4y)\) Simplifying: \(x + \frac{dy}{dx}x + 4y^2 + 4xy\frac{dy}{dx} = 1\) Finally, isolating \(\frac{dy}{dx}\) yields: \(\frac{dy}{dx} = \frac{4y^2 - x}{1-4xy}\) 6. **Find \(\frac{d^2y}{dx^2}\)**: Differentiate \(\frac{dy}{dx} = \frac{4y^2 - x}{1-4xy}\) implicitly: Using quotient rule: \(\frac{d^2y}{dx^2} = \frac{(1-4xy)\left( \frac{d}{dx}(4y^2-x) \right) - (4y^2-x) \left( \frac{d}{dx}(1-4xy) \right)}{(1-4xy)^2}\) The derivatives are: \(\frac{d}{dx}(4y^2-x) = 8y\frac{dy}{dx}-1\) and \(\frac{d}{dx}(1-4xy) = -4y-4x\frac{dy}{dx}\) 7. **Set \(x=0\) and evaluate \(y\):** From \(\log_e(x+y)=4xy\) at \(x=0\), \(\log_e(y)=0\) implies \(y=1\). At \(x=0\), \(\frac{dy}{dx} = \frac{4(1)^2-0}{1-4\cdot0\cdot1} = 4.\) 8. **Substitute for \(\frac{d^2y}{dx^2}\) at \(x=0\), \(y=1\):** \(\frac{d^2y}{dx^2} = \frac{(1-0)(8\cdot1\cdot4-1) - (4-0)(-4\cdot1)}{(1)^2}\) Simplify: \(\frac{d^2y}{dx^2} = \frac{-1 + 16 + 16}{1} = \frac{31}{1} = 31\). **Conclusion**: \(\frac{d^2y}{dx^2}\) at \(x=0\) is 31, which lies within the given range (40, 40).
