If $x=\log t, t>0$ and $\mathrm{y}=\frac{1}{\mathrm{t}}$ then $\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{d} x^2}=$

Given the parametric equations $x = \log t$ and $y = \frac{1}{t}$, the first derivative is determined using the chain rule:

$\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{\frac{\mathrm{d} y}{\mathrm{d} t}}{\frac{\mathrm{d} x}{\mathrm{d} t}} = \frac{-\frac{1}{t^2}}{\frac{1}{t}} = -\frac{1}{t}$

For the second derivative, $\frac{\mathrm{d}^2 y}{\mathrm{d} x^2} = \frac{\mathrm{d}}{\mathrm{d} t} \left( \frac{\mathrm{d} y}{\mathrm{d} x} \right) \cdot \frac{\mathrm{d} t}{\mathrm{d} x}$. Differentiating $\frac{\mathrm{d} y}{\mathrm{d} x} = -\frac{1}{t}$ with respect to $t$ yields $\frac{1}{t^2}$, and since $\frac{\mathrm{d} x}{\mathrm{d} t} = \frac{1}{t}$, it follows that $\frac{\mathrm{d} t}{\mathrm{d} x} = t$. Consequently:

$\frac{\mathrm{d}^2 y}{\mathrm{d} x^2} = \frac{1}{t^2} \cdot t = \frac{1}{t}$

Observing that $\frac{1}{t} = -\left( -\frac{1}{t} \right) = -\frac{\mathrm{d} y}{\mathrm{d} x}$, the result simplifies to $\frac{\mathrm{d}^2 y}{\mathrm{d} x^2} = -\frac{\mathrm{d} y}{\mathrm{d} x}$, which corresponds to option B.