Find dy/dx, if x=log⁡t² and y=1/t

The correct option is – (1/2t).

To find $\frac{dy}{dx}$ when $x$ and $y$ are given as functions of a parameter $t$, we use the chain rule for parametric differentiation:
$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

First, let's simplify and differentiate $x$ with respect to $t$:
Given, $x=\log \left(t^2\right)$.
Using the logarithm property $\log \left(a^b\right)=b\log \left(a\right)$, we can rewrite $x$ as:
$x=2\log \left(t\right)$.
Now, differentiating $x$ with respect to $t$:
$\frac{dx}{dt}=2\cdot \frac{1}{t}=\frac{2}{t}$.

Next, let's differentiate $y$ with respect to $t$:
Given, $y=\frac{1}{t}=t^{-1}$.
Differentiating $y$ with respect to $t$ using the power rule:
$\frac{dy}{dt}=-1\cdot t^{-2}=-\frac{1}{t^2}$.

Finally, we substitute these derivatives into the parametric differentiation formula:
$\frac{dy}{dx}=\frac{-\frac{1}{t^2}}{\frac{2}{t}}$
Simplifying the expression:
$\frac{dy}{dx}=-\frac{1}{t^2}\cdot \frac{t}{2}=-\frac{1}{2t}$.

Thus, $\frac{dy}{dx}=–(1/2t)$.