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CONTINUITY AND DIFFERENTIABILITY — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSCONTINUITY AND DIFFERENTIABILITY3 Marks
Question
If $\text{x}=\text{e}^{\frac{\text{x}}{\text{y}}},$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\text{x}-\text{y}}{\text{x}\log\text{x}}.$
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Answer

We have, $\text{x}=\text{e}^{\frac{\text{x}}{\text{y}}}$
Differentiating both sides w.r.t. x, we get
$\therefore\ 1=\text{e}^{\frac{\text{x}}{\text{y}}}\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}}{\text{y}}\Big)$
$\Rightarrow\ 1=\text{e}^{\frac{\text{x}}{\text{y}}}\bigg[\frac{\text{y}\cdot1-\text{x}\frac{\text{dy}}{\text{dx}}}{\text{y}^2}\bigg]$
$\Rightarrow\ \text{y}^2=\text{y}\cdot\text{e}^{\frac{\text{x}}{\text{y}}}-\text{x}\cdot\frac{\text{dy}}{\text{dx}}\cdot\text{e}^{\frac{\text{x}}{\text{y}}}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{y}\Big(\text{e}^{\frac{\text{x}}{\text{y}}}-\text{y}\Big)}{\text{x}\cdot\text{e}^{\frac{\text{x}}{\text{y}}}}$
$=\frac{\text{e}^{\frac{\text{x}}{\text{y}}}-\text{y}}{\frac{\frac{\text{x}}{\text{y}}\text{e}^{\frac{\text{x}}{\text{y}}}}{}}$
$=\frac{\text{x}-\text{y}}{\text{x}\cdot\log\text{x}}$ $\Big[\because\ \text{x}=\text{e}^{\frac{\text{x}}{\text{y}}}\Rightarrow\log\text{x}=\frac{\text{x}}{\text{y}}\Big]$
Hence proved

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