MCQ
જો $y = \sqrt {\sin x + y} ,$ તો ${{dy} \over {dx}}$ મેળવો.
- A${{\sin x} \over {2y - 1}}$
- ✓${{\cos x} \over {2y - 1}}$
- C${{\sin x} \over {2y + 1}}$
- D${{\cos x} \over {2y + 1}}$
Differentiate with respect to $ x$ ,
$2y.\frac{{dy}}{{dx}} = \cos x + \frac{{dy}}{{dx}}$
==> $\frac{{dy}}{{dx}}(2y - 1) = \cos x$
==> $\frac{{dy}}{{dx}} = \frac{{\cos x}}{{2y - 1}}$.
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