A) \[\sin (\sin {{x}^{2}}).\cos {{x}^{2}}.2x\]
B) \[-\sin (\sin {{x}^{2}}).\cos {{x}^{2}}.2x\]
C) \[-\sin (\sin {{x}^{2}}).{{\cos }^{2}}x.2x\]
D) None of these
Correct Answer: B
Solution :
\[\frac{d}{dx}\{\cos (\sin {{x}^{2}})\}=-\sin (\sin {{x}^{2}})\cos {{x}^{2}}.2x\].
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