If (xy) + x y = x^2 - y, then dy dx = - Vidyadip

MCQ

If $\sin (xy) + {x \over y} = {x^2} - y,$ then ${{dy} \over {dx}} = $

✓
${{y[2xy - {y^2}\cos (xy) - 1]} \over {x{y^2}\cos (xy) + {y^2} - x}}$
B
${{[2xy - {y^2}\cos (xy) - 1]} \over {x{y^2}\cos (xy) + {y^2} - x}}$
C
$ - {{y[2xy - {y^2}\cos (xy) - 1]} \over {x{y^2}\cos (xy) + {y^2} - x}}$
D
None of these

✓

Answer

Correct option: A.

${{y[2xy - {y^2}\cos (xy) - 1]} \over {x{y^2}\cos (xy) + {y^2} - x}}$

a
(a) $\sin (xy) + \frac{x}{y} = {x^2} - y$

Differentiating both sides,

$\cos (xy)\frac{d}{{dx}}(xy) + x\left\{ { - \frac{1}{{{y^2}}}} \right\}\frac{{dy}}{{dx}} + \frac{1}{y} = 2x - \frac{{dy}}{{dx}}$

==> $[x\cos (xy) - \frac{x}{{{y^2}}} + 1]\frac{{dy}}{{dx}} = 2x - \frac{1}{y} - y\cos (xy)$

==> $\frac{{dy}}{{dx}} = \left[ {\frac{{2x{y^2} - y - {y^3}\cos (xy)}}{{x{y^2}\cos (xy) - x + {y^2}}}} \right]$.

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