(115) Ify \( =\ln 4 t \) and \( x=\ln (2 t+1) \), the \( \frac{d^{2} y}{d x^{2}}=\cdots \) at \( t=1 \) \( \begin{array}{llll}\text { a) }-\frac{4}{3} & \text { b) }-\frac{3}{4} & \text { c) }-4 & \text { d) }-\frac{1}{2}\end{array} \)

To find \(\frac{d^2 y}{dx^2}\) at \(t=1\) given \(y = \ln(4t)\) and \(x = \ln(2t + 1)\), we will first compute \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\), and then use the chain rule to find \(\frac{dy}{dx}\) and \(\frac{d^2y}{dx^2}\). 1. **Calculate \(\frac{dy}{dt}\)**: \[ y = \ln(4t) = \ln(4) + \ln(t) \] Therefore, \[ \frac{dy}{dt} = \frac{1}{t} \] 2. **Calculate \(\frac{dx}{dt}\)**: \[ x = \ln(2t + 1) \] Using the chain rule, \[ \frac{dx}{dt} = \frac{2}{2t + 1} \] 3. **Calculate \(\frac{dy}{dx}\)**: Using the chain rule, \[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{1}{t}}{\frac{2}{2t + 1}} = \frac{2(2t + 1)}{2t} = \frac{2(2t + 1)}{2t} = \frac{2t + 1}{t} \] 4. **Calculate \(\frac{d^2y}{dx^2}\)**: We need to differentiate \(\frac{dy}{dx}\) with respect to \(t\) and then divide by \(\frac{dx}{dt}\): \[ \frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{2t + 1}{t}\right) \] Using the quotient rule: \[ \frac{d}{dt}\left(\frac{2t + 1}{t}\right) = \frac{(t)(2) - (2t + 1)(1)}{t^2} = \frac{2t - (2t + 1)}{t^2} = \frac{-1}{t^2} \] Now, we can find \(\frac{d^2y}{dx^2}\): \[ \frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} = \frac{\frac{-1}{t^2}}{\frac{2}{2t + 1}} = \frac{-1}{t^2} \cdot \frac{2t + 1}{2} = \frac{-2(2t + 1)}{2t^2} = \frac{-(2t + 1)}{t^2} \] 5. **Evaluate at \(t = 1\)**: \[ \frac{d^2y}{dx^2} \bigg|_{t=1} = \frac{-(2(1) + 1)}{(1)^2} = \frac{-3}{1} = -3 \] Thus, the final answer is: \[ \frac{d^2y}{dx^2} = -3 \] Since this value does not match any of the provided options, it seems there may have been a misunderstanding in the options or the calculations. However, based on the calculations, the answer is \(-3\).