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}} $
  • $0$

Answer

Correct option: D.
$0$
d
(d) Since $\frac{{dy}}{{dx}} = - \sin (\sin {x^2}).\cos {x^2}.2x$

Therefore, at $x = \sqrt {\frac{\pi }{2}} ,$ $\cos {x^2} = \cos \frac{\pi }{2} = 0$

$\Rightarrow $ $\frac{{dy}}{{dx}} = 0$.

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