Answer is: option1
\( -\frac{1}{4} \)Solution:
\( x = t^2 \quad \Rightarrow \quad \frac{dx}{dt} = 2t \)
\( y = \ln(t^2 + 1) \quad \Rightarrow \quad \frac{dy}{dt} = \frac{1}{t^2 + 1} \cdot 2t \)
\( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{1}{t^2 + 1} \)
\( \frac{d^2 y}{dx^2} = \frac{d}{dt} \left( \frac{dy}{dx} \right) \div \frac{dx}{dt} \)
\( = \frac{-2t}{(t^2 + 1)^2} \div 2t \)
\( = \frac{-1}{(t^2 + 1)^2} \)
Then, \( \left. \frac{d^2 y}{dx^2} \right|_{t=1} = \frac{-1}{(t^2 + 1)^2} \Big|_{t=1} = \frac{-1}{4} \)
