MCQ
જો $\sin y + {e^{ - x\,\cos y}} = e,$ તો ${{dy} \over {dx}}$ એ $(1,\pi )$ આગળ મેળવો.
- A$\sin y$
- B$ - x\cos y$
- ✓$e$
- D$\sin y - x\,\cos y$
==> $\cos y\frac{{dy}}{{dx}} + {e^{ - x\cos y}}\left\{ {( - x)\,\left( { - \sin y\frac{{dy}}{{dx}}} \right) + \cos y( - 1)} \right\}\, = 0$
==> $\cos y\frac{{dy}}{{dx}} + x\sin y\,\,{e^{ - x\cos y}}\frac{{dy}}{{dx}} - \cos y{e^{ - x\cos y}} = 0$
==> $\frac{{dy}}{{dx}} = \frac{{\cos y\,\,{e^{ - x\cos y}}}}{{\cos y + x\sin y\,\,{e^{ - x\cos y}}}}$
==> ${\left( {\frac{{dy}}{{dx}}} \right)_{(1,\,\pi )}} = \frac{{\cos \pi \,\,{e^{ - \cos \pi }}}}{{\cos \pi + \sin \pi \,\,{e^{ - \cos \pi }}}}$
$= \frac{{( - 1)e}}{{ - 1 + 0}} = e$.
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