Continuity and Differentiability — Class 12 Maths Solution

We have, ${\left( {{x^2} + {y^2}} \right)^2} = xy$

On differentiating both sides w.r.t. $x$, we get
$\frac{d}{{dx}}{\left( {{x^2} + {y^2}} \right)^2} = \frac{d}{{dx}}(xy)$

$\Rightarrow$ $2\left( {{x^2} + {y^2}} \right) \cdot \frac{d}{{dx}}\left( {{x^2} + {y^2}} \right) = x \cdot \frac{d}{{dx}}y + y \cdot \frac{d}{{dx}}x$

$\Rightarrow$ $2\left( {{x^2} + {y^2}} \right) \cdot \left( {2x + 2y\frac{{dy}}{{dx}}} \right) = x\frac{{dy}}{{dx}} + y$

$\Rightarrow$ $2{x^2} \cdot 2x + 2{x^2} \cdot 2y\frac{{dy}}{{dx}} + 2{y^2} \cdot 2x + 2{y^2} \cdot 2y\frac{{dy}}{{dx}} = x\frac{{dy}}{{dx}} + y$

$\Rightarrow$ $\frac{{dy}}{{dx}}\left[ {4{x^2}y + 4{y^3} - x} \right] = y - 4{x^3} - 4x{y^2}$

therefore,$\frac{{dy}}{{dx}} = \frac{{\left( {y - 4{x^3} - 4x{y^2}} \right)}}{{\left( {4{x^2}y + 4{y^3} - x} \right)}}$