d dx (x^2) = - Vidyadip

MCQ

${d \over {dx}}\{ \cos (\sin {x^2})\} = $

A
$\sin (\sin {x^2}).\cos {x^2}.2x$
✓
$ - \sin (\sin {x^2}).\cos {x^2}.2x$
C
$ - \sin (\sin {x^2}).{\cos ^2}x.2x$
D
None of these

✓

Answer

Correct option: B.

$ - \sin (\sin {x^2}).\cos {x^2}.2x$

b
(b) $\frac{d}{{dx}}\{ \cos (\sin {x^2})\} = - \sin (\sin {x^2})\cos {x^2}.2x$.

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