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CONTINUITY AND DIFFERENTIABILITY — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Find $\frac{\text{dy}}{\text{dx}}$ of the functions expressed in parametric:
$\text{x}=\frac{1+\log\text{t}}{\text{t}^2},\text{ y}=\frac{3+2\log\text{t}}{\text{t}}.$
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Answer

Consider, $\text{x}=\frac{1+\log\text{t}}{\text{t}^2}$ and $\text{y}=\frac{3+2\log\text{t}}{\text{t}}$
$\Rightarrow\ \frac{\text{dx}}{\text{dt}}=\frac{\text{t}^2\cdot\frac{\text{d}}{\text{dt}}(1+\log\text{t})-(1+\log\text{t})\cdot\frac{\text{d}}{\text{dt}}\text{t}^2}{(\text{t}^2)^2}$
$=\frac{\text{t}^2\cdot\frac{1}{\text{t}}-(1+\log\text{t})\cdot2\text{t}}{\text{t}^4}$
$=\frac{\text{t}-(1+\log\text{t})\cdot2\text{t}}{\text{t}^4}$
$=\frac{-1-2\log\text{t}}{\text{t}^3}$
and $\frac{\text{dy}}{\text{dt}}=\frac{\text{t}\cdot\frac{\text{d}}{\text{dt}}(3+2\log\text{t})-(3+2\log\text{t})\cdot\frac{\text{d}}{\text{dt}}\text{t}}{\text{t}^2}$
$=\frac{\text{t}\cdot2\cdot\frac{1}{\text{t}}-(3+2\log\text{t})\cdot1}{\text{t}^2}$a
$=\frac{-1-2\log\text{t}}{\text{t}^2}$
$\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{-1-2\log\frac{\text{t}}{\text{t}^2}}{-1-2\log\frac{\text{t}}{\text{t}^3}}=\text{t}$

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