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Mathematics · 2023 · 2 marks
CBSE 2023 · Region 4 · Set 1 · Q22
If ( x^2+ y^2)^2= x y, then find (d y)/(d x).
If $\displaystyle \left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)^{2}=\mathrm{x} \mathrm{y}$, then find $\displaystyle \frac{d \mathrm{y}}{d \mathrm{x}}$.
Marking-scheme solution
$$\begin{gathered}
\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)^{2}=\mathrm{xy} \text { gives } \\
2\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)\left[\mathbf{2} \boldsymbol{\mathrm{x}}+\mathbf{2} \boldsymbol{\mathrm{y}} \frac{\boldsymbol{d} \boldsymbol{\mathrm{y}}}{\boldsymbol{d} \boldsymbol{\mathrm{x}}}\right]=\boldsymbol{\mathrm{x}} \frac{\boldsymbol{d} \boldsymbol{\mathrm{y}}}{\boldsymbol{d} \boldsymbol{\mathrm{x}}}+\boldsymbol{\mathrm{y}} \\
\Rightarrow\left[4 \mathrm{y}\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)-\mathrm{x}\right] \frac{\boldsymbol{d} \boldsymbol{\mathrm{y}}}{\boldsymbol{d} \boldsymbol{\mathrm{x}}}=\mathrm{y}-4 \mathrm{x}\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right) \\
\Rightarrow \frac{\boldsymbol{d} \boldsymbol{\mathrm{y}}}{\boldsymbol{d} \boldsymbol{\mathrm{x}}}=\frac{\boldsymbol{\mathrm{y}}-\mathbf{4} \boldsymbol{\mathrm{x}}\left(\boldsymbol{\mathrm{x}}^{2}+\boldsymbol{\mathrm{y}}^{2}\right)}{\mathbf{4} \boldsymbol{\mathrm{y}}\left(\boldsymbol{\mathrm{x}}^{2}+\boldsymbol{\mathrm{y}}^{2}\right)-\boldsymbol{\mathrm{x}}}
\end{gathered}
$$
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CBSE Class 12 Mathematics past-paper question from the 2023 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.