MCQ
જો $x{e^{xy}} = y + {\sin ^2}x$ ,તો $x = 0,{{dy} \over {dx}} = $
- A$-1$
- B$-2$
- ✓$1$
- D$2$
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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