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: Conceptualization:
The objective is to derive a 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 then substituting these into the given options to identify the correct relationship.

Step 2: Derivation:
Given equation:
\[ e^y = \log x \]
First Derivative:
Implicitly differentiate both sides with respect to x:
\[ \frac{d}{dx}(e^y) = \frac{d}{dx}(\log x) \]
Applying the chain rule to the left side:
\[ e^y \frac{dy}{dx} = \frac{1}{x} \]
Rearranging yields:
\[ x e^y \frac{dy}{dx} = 1 \quad \cdots (1) \]

Second Derivative:
Differentiate equation (1) with respect to x using the product rule for three functions (u.v.w)' = u'vw + uv'w + uvw' where u=x, v=$e^y$, w=dy/dx.
\[ \frac{d}{dx}\left(x e^y \frac{dy}{dx}\right) = \frac{d}{dx}(1) \]
\[ \left(\frac{d}{dx}x\right) \left(e^y \frac{dy}{dx}\right) + x \left(\frac{d}{dx}e^y\right) \left(\frac{dy}{dx}\right) + x e^y \left(\frac{d}{dx}\frac{dy}{dx}\right) = 0 \]
\[ (1) \left(e^y \frac{dy}{dx}\right) + x \left(e^y \frac{dy}{dx}\right) \left(\frac{dy}{dx}\right) + x e^y \left(\frac{d^2y}{dx^2}\right) = 0 \]
\[ e^y \frac{dy}{dx} + x e^y \left(\frac{dy}{dx}\right)^2 + x e^y \frac{d^2y}{dx^2} = 0 \]
As \(e^y\) is always positive, divide the equation by \(e^y\):
\[ \frac{dy}{dx} + x \left(\frac{dy}{dx}\right)^2 + x \frac{d^2y}{dx^2} = 0 \]

Step 3: Conclusion:
Rearrange 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 equation corresponds to option (D).