If \(e^y = \log x\), then which of the following is true?
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When performing implicit differentiation a second time, it's often easier to differentiate a simplified form of the first derivative. Here, differentiating \(x e^y \frac{dy}{dx} = 1\) is simpler than differentiating \(\frac{dy}{dx} = \frac{1}{x e^y}\) which would require the quotient rule and lead to more complex substitutions.
Step 1: Concept Identification: The objective is to determine the differential equation satisfied by the provided function. This necessitates calculating the first and second derivatives of y with respect to x via implicit differentiation and subsequently substituting these into the given choices to identify the correct relationship.
Step 2: Derivation Process: The given equation is:\[ e^y = \log x \] First Derivative Calculation: Differentiating both sides with respect to x: \[ \frac{d}{dx}(e^y) = \frac{d}{dx}(\log x) \] Applying the chain rule to the left side yields: \[ e^y \frac{dy}{dx} = \frac{1}{x} \] Rearranging this equation results in: \[ x e^y \frac{dy}{dx} = 1 \quad \cdots (1) \]
Second Derivative Calculation: Differentiating equation (1) with respect to x. The product rule for three functions (u.v.w)' = u'vw + uv'w + uvw' is applied to the left side, with u=x, v=$e^y$, and w=dy/dx. \[ \frac{d}{dx}\left(x e^y \frac{dy}{dx}\right) = \frac{d}{dx}(1) \] \[ \left(\frac{d}{dx}x\right) \left(e^y \frac{dy}{dx}\right) + x \left(\frac{d}{dx}e^y\right) \left(\frac{dy}{dx}\right) + x e^y \left(\frac{d}{dx}\frac{dy}{dx}\right) = 0 \] \[ (1) \left(e^y \frac{dy}{dx}\right) + x \left(e^y \frac{dy}{dx}\right) \left(\frac{dy}{dx}\right) + x e^y \left(\frac{d^2y}{dx^2}\right) = 0 \] \[ e^y \frac{dy}{dx} + x e^y \left(\frac{dy}{dx}\right)^2 + x e^y \frac{d^2y}{dx^2} = 0 \] As \(e^y\) is consistently positive, the equation can be divided by \(e^y\): \[ \frac{dy}{dx} + x \left(\frac{dy}{dx}\right)^2 + x \frac{d^2y}{dx^2} = 0 \]
Step 3: Final Solution: Rearranging the terms to align with the provided options: \[ x \frac{d^2y}{dx^2} + x \left(\frac{dy}{dx}\right)^2 + \frac{dy}{dx} = 0 \] This matches option (D).