A) m
B) \[(2m)!\]
C) 2m
D) m!
Correct Answer: B
Solution :
Expanding binomially \[y={{({{x}^{2}}-1)}^{m}}{{=}^{m}}{{C}_{o}}{{x}^{2m}}{{+}^{m}}{{C}_{1}}{{x}^{2m-2}}(-1)+.......\] So on differentiating all the terms except first reduces to zero, therefore \[\frac{{{d}^{2m}}y}{d{{x}^{2m}}}{{=}^{m}}{{C}_{0}}2m\text{ }(2m-1)\text{ }(2m-2)....1=(2m)!\].
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