MCQ
If $y = \cos (\sin {x^2}),$ then at $x = \sqrt {{\pi \over 2}} ,{{dy} \over {dx}}=$
- A$-2$
- B$2$
- C$ - 2\sqrt {{\pi \over 2}} $
- ✓$0$
Therefore, at $x = \sqrt {\frac{\pi }{2}} ,$ $\cos {x^2} = \cos \frac{\pi }{2} = 0$
$\Rightarrow $ $\frac{{dy}}{{dx}} = 0$.
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