Question
If $x{e^{xy}} = y + {\sin ^2}x$, then at $x = 0,{{dy} \over {dx}} = $
✓
Answer
c
(c) We are given that $x{e^{xy}} = y + {\sin ^2}x$
(c) We are given that $x{e^{xy}} = y + {\sin ^2}x$
When $x = 0$, we get $y = 0$
Differentiating both sides with respect to $x$, we get
${e^{xy}} + x{e^{xy}}\left[ {x\frac{{dy}}{{dx}} + y} \right] = \frac{{dy}}{{dx}} + 2\sin x\cos x$
Putting $x = 0,\,y = 0$, we get $\frac{{dy}}{{dx}} = 1$.
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