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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY3 Marks
Question
Find $\frac{\text{dy}}{\text{dx}}$ when x and y are connected by the relation:
$\big(\text{x}^2+\text{y}^2\big)^2=\text{xy}$
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Answer

We have, $\big(\text{x}^2+\text{y}^2\big)^2=\text{xy}$
On differentiating both sides w.r.t. x, we get
$\frac{\text{d}}{\text{dx}}\big(\text{x}^2+\text{y}^2\big)^2=\frac{\text{d}}{\text{dx}}(\text{xy})$
$\Rightarrow\ 2\big(\text{x}^2+\text{y}^2\big)^2\cdot\frac{\text{d}}{\text{dx}}(\text{x}^2+\text{y}^2)=\text{x}\cdot\frac{\text{d}}{\text{dx}}\text{y}+\text{y}\cdot\frac{\text{d}}{\text{dx}}\text{x}$
$\Rightarrow\ 2(\text{x}^2+\text{y}^2)\cdot\Big(2\text{x}+2\text{y}\frac{\text{dy}}{\text{dy}}\Big)=\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}$
$\Rightarrow\ 2\text{x}^2\cdot2\text{x}+2\text{x}^2\cdot2\text{y}\frac{\text{dy}}{\text{dx}}+2\text{y}^2\cdot2\text{x}+2\text{y}^2\cdot2\text{y}\frac{\text{dy}}{\text{dx}}=\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}\big[4\text{x}^2\text{y}+4\text{y}^3-\text{x}\big]=\text{y}-4\text{x}^3-4\text{xy}^2$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{(\text{y}-4\text{x}^3-4\text{xy}^2)}{(4\text{x}^2\text{y}+4\text{y}^3-\text{x})}$

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