If \(x = \displaystyle\frac{1 + \log_e t}{t^{2}}\) and \(y = \displaystyle\frac{3 + 2\log_e t}{t}\), where t > 0 and \(1 + 2\log_e t \ne 0\), then \(\displaystyle\frac{dy}{dx}\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(t\)
By the quotient rule:
\( \displaystyle \frac{dx}{dt} = \frac{\frac{1}{t} \cdot t^{2} - (1 + \log_e t) \cdot 2t}{t^{4}} \)
\( \displaystyle = \frac{t - 2t - 2t\log_e t}{t^{4}} = -\frac{1 + 2\log_e t}{t^{3}} \)
\( \displaystyle \frac{dy}{dt} = \frac{\frac{2}{t} \cdot t - (3 + 2\log_e t)}{t^{2}} \)
\( \displaystyle = -\frac{1 + 2\log_e t}{t^{2}} \)
In parametric form:
\( \displaystyle \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{t^{3}}{t^{2}} = t \)