Question
यदि $\sin (xy) + \frac{x}{y} = {x^2} - y,$ तो $\frac{{dy}}{{dx}} = $
दोनों पक्षों का $x$ के सापेक्ष अवकलन करने पर,
$\cos (xy)\frac{d}{{dx}}(xy) + x\left\{ { - \frac{1}{{{y^2}}}} \right\}\frac{{dy}}{{dx}} + \frac{1}{y} = 2x - \frac{{dy}}{{dx}}$
==> $[x\cos (xy) - \frac{x}{{{y^2}}} + 1]\frac{{dy}}{{dx}} = 2x - \frac{1}{y} - y\cos (xy)$
==> $\frac{{dy}}{{dx}} = \left[ {\frac{{2x{y^2} - y - {y^3}\cos (xy)}}{{x{y^2}\cos (xy) - x + {y^2}}}} \right]$.
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