Question

If \(e^y = \log x\), then which of the following is true?

• A. \(x\frac{d^2y}{dx^2} - \left(\frac{dy}{dx}\right)^2 + \frac{dy}{dx} = 0\)
• B. \(\frac{d^2y}{dx^2} - x\frac{dy}{dx} = 0\)
• C. \(\frac{d^2y}{dx^2} - \left(\frac{dy}{dx}\right)^2 + 1 = 0\)
• D. \(x\frac{d^2y}{dx^2} + x\left(\frac{dy}{dx}\right)^2 + \frac{dy}{dx} = 0\)

Hint:

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.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The problem requires finding a differential equation that is satisfied by the given function. This involves finding the first and second derivatives of y with respect to x using implicit differentiation and then substituting them into the given options to find the correct relationship.
Step 2: Detailed Explanation:
Given the equation: \[ e^y = \log x \] First Derivative: Differentiate both sides with respect to x: \[ \frac{d}{dx}(e^y) = \frac{d}{dx}(\log x) \] Using the chain rule on the left side: \[ e^y \frac{dy}{dx} = \frac{1}{x} \] Rearranging this equation, we get: \[ x e^y \frac{dy}{dx} = 1 \quad \cdots (1) \] Second Derivative: Now, differentiate equation (1) with respect to x. We use the product rule for three functions (u.v.w)' = u'vw + uv'w + uvw' on the left side, where u=x, v=$e^y$, w=dy/dx. \[ \frac{d}{dx}\left(x e^y \frac{dy}{dx}\right) = \frac{d}{dx}(1) \] \[ \left(\frac{d}{dx}x\right) \left(e^y \frac{dy}{dx}\right) + x \left(\frac{d}{dx}e^y\right) \left(\frac{dy}{dx}\right) + x e^y \left(\frac{d}{dx}\frac{dy}{dx}\right) = 0 \] \[ (1) \left(e^y \frac{dy}{dx}\right) + x \left(e^y \frac{dy}{dx}\right) \left(\frac{dy}{dx}\right) + x e^y \left(\frac{d^2y}{dx^2}\right) = 0 \] \[ e^y \frac{dy}{dx} + x e^y \left(\frac{dy}{dx}\right)^2 + x e^y \frac{d^2y}{dx^2} = 0 \] Since \(e^y\) is always positive, we can divide the entire equation by \(e^y\): \[ \frac{dy}{dx} + x \left(\frac{dy}{dx}\right)^2 + x \frac{d^2y}{dx^2} = 0 \] Step 3: Final Answer:
Rearranging the terms to match the options, we get: \[ x \frac{d^2y}{dx^2} + x \left(\frac{dy}{dx}\right)^2 + \frac{dy}{dx} = 0 \] This matches option (D).