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Question:

Find $\frac{d y}{d x}$, when

If $x=\frac{1+\log t}{t^{2}}, y=\frac{3+2 \log t}{t}$, find $\frac{d y}{d x}$

Solution:

$: x=\frac{1+\log t}{t^{2}}, y=\frac{3+2 \log t}{t}$

$\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\mathrm{t}^{2}\left(\frac{1}{\mathrm{t}}\right)-(1+\log \mathrm{t})(2 \mathrm{t})}{\mathrm{t}^{4}}=\frac{\mathrm{t}-2 \mathrm{t}-2 \mathrm{t} \log \mathrm{t}}{\mathrm{t}^{4}}=\frac{-2 \log \mathrm{t}-1}{\mathrm{t}^{3}}$

$\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{\mathrm{t}\left(\frac{2}{\mathrm{t}}\right)-(3+2 \log \mathrm{t})(1)}{\mathrm{t}^{2}}=\frac{2-3-2 \log \mathrm{t}}{\mathrm{t}^{2}}=\frac{-2 \log \mathrm{t}-1}{\mathrm{t}^{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}=\frac{\frac{-2 \log \mathrm{t}-1}{\mathrm{t}^{2}}}{\frac{-2 \log \mathrm{t}-1}{\mathrm{t}^{3}}}=\mathrm{t}$

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