Q.If $x = \log(1 + t^2)$ and $y = \log t$, then find $\dfrac{dy}{dx}$.

Differentiate the parametric forms: $\frac{dx}{dt} = \frac{2t}{1+t^2}$, $\frac{dy}{dt} = \frac{1}{t}$, so $\frac{dy}{dx} = \frac{1+t^2}{2t^2}$.

Differentiate each equation with respect to $t$.

From $x = \log(1 + t^2)$:

$\frac{dx}{dt} = \frac{1}{1 + t^2}\cdot 2t = \frac{2t}{1 + t^2}.$

From $y = \log t$:

$\frac{dy}{dt} = \frac{1}{t}.$

Therefore …