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Question
If \[x = \frac{1 + \log t}{t^2}, y = \frac{3 + 2\log t}{t}, \text { find } \frac{dy}{dx}\] ?
Solution
\[x = \frac{1 + \log t}{t^2} \text { and y } = \frac{3 + 2\log t}{t}\]
\[ \Rightarrow \frac{dx}{dt} = \frac{t - 2t - 2t\log t}{t^4} \text { and } \frac{dy}{dt} = \frac{2 - 3 - 2\log t}{t^2}\]
\[ \Rightarrow \frac{dx}{dt} = \frac{- 1 - 2\log t}{t^3} \text { and } \frac{dy}{dt} = \frac{- 1 - 2\log t}{t^2}\]
\[ \therefore \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{- 1 - 2\log t}{t^2}}{\frac{- 1 - 2\log t}{t^3}} = t\]
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