Find $\frac{dy}{dx}$ if $x = \frac{1 + \log t}{t^2}$ and $y = \frac{3 + 2 \log t}{t}$.

$t$

The correct answer is Option (2) → $t$ ##

We have, $x = \frac{1 + \log t}{t^2}$ and $y = \frac{3 + 2 \log t}{t}$.

Taking derivative of both equations w.r.t. $t$, we get

$∴\frac{dx}{dt} = \frac{t^2 \cdot \frac{d}{dt}(1 + \log t) - (1 + \log t) \cdot \frac{d}{dt}t^2}{(t^2)^2} \quad \text{[by quotient rule]}$

$= \frac{t^2 \cdot \frac{1}{t} - (1 + \log t) \cdot 2t}{t^4} = \frac{t - (1 + \log t) \cdot 2t}{t^4}$

$= \frac{t}{t^4} [1 - 2(1 + \log t)] = \frac{-1 - 2 \log t}{t^3} \dots (i)$

and, $\frac{dy}{dt} = \frac{t \cdot \frac{d}{dt}(3 + 2 \log t) - (3 + 2 \log t) \cdot \frac{d}{dt}t}{t^2}$

$= \frac{t \cdot 2 \cdot \frac{1}{t} - (3 + 2 \log t) \cdot 1}{t^2}$

$= \frac{2 - 3 - 2 \log t}{t^2} = \frac{-1 - 2 \log t}{t^2} \dots (ii)$

On dividing Eq. (ii) by Eq. (i), we get

$∴\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{(-1 - 2 \log t) / t^2}{(-1 - 2 \log t) / t^3} = t$