Question

If \(x = e^{y+e^y+.... to \space ∞}, x> 0\) then \(\frac{d^2y}{dx^2}\) is:

• A. \(-x^2\)
• B. \(-\frac{1}{x^2}\)
• C. \(\frac{1}{x^2}\)
• D. \(x^2\)

Answer 1

The Correct Option is B

To solve the problem, we start with the given equation \(x = e^{y+e^y+\ldots \text{to} \, \infty}\). This equation can be simplified by noting the recursive nature where \(y = y + e^y + e^{e^y} + \ldots\). Hence, \(x = e^{y+e^y+\ldots} = e^y\). Thus, we can express the relationship as \(x = e^y\), implying \(y = \ln x\).

Now, let's find the derivative of \(y\) with respect to \(x\):

\(\frac{dy}{dx} = \frac{d}{dx}(\ln x) = \frac{1}{x}\)

Next, we need to find the second derivative \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{1}{x}\right) = \frac{d}{dx}(x^{-1})\)

Using the power rule, the derivative of \(x^{-1}\) is:

\(-x^{-2} = -\frac{1}{x^2}\)

Therefore, the second derivative \(\frac{d^2y}{dx^2}\) is:

\(-\frac{1}{x^2}\)

This result matches the correct answer from the provided options.