A) \[\frac{1}{2}x\cos {{x}^{3}}\cos ec\,{{x}^{2}}\]
B) \[\frac{3}{2}x\cos {{x}^{3}}\cos ec\,{{x}^{2}}\]
C) \[\frac{1}{2}x\sec {{x}^{3}}\sin \,{{x}^{2}}\]
D) \[\frac{3}{2}x\sec {{x}^{3}}\cos ec\,{{x}^{2}}\]
Correct Answer: B
Solution :
[b] Here, \[u=f({{x}^{3}})\] \[\Rightarrow \frac{du}{dx}=f'({{x}^{3}}).\frac{d}{dx}({{x}^{3}})\] \[=(\cos ({{x}^{3}})).3{{x}^{2}}=3{{x}^{2}}.\cos {{x}^{3}}\] and \[v=g({{x}^{2}})\] \[\Rightarrow \frac{dv}{dx}=g'({{x}^{2}}).\frac{d}{dx}({{x}^{2}})=(\sin {{x}^{2}}).(2x)\] \[=2x\sin {{x}^{2}}\] \[\therefore \frac{du}{dv}=\frac{\frac{du}{dx}}{\frac{dv}{dx}}\frac{3{{x}^{2}}.\cos {{x}^{2}}}{2x.\sin {{x}^{2}}}\] \[\Rightarrow \frac{du}{dv}=\frac{3}{2}x.\cos {{x}^{3}}.\cos ec\,{{x}^{2}}\]
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