If \[u=f({{x}^{3}}),v=g({{x}^{2}}),f'(x)=\cos x\] and \[g'(x)=\sin x,\] then \[\frac{du}{dv}=\]

\[\frac{1}{2}x\sec {{x}^{3}}\sin \,{{x}^{2}}\]

\[\frac{1}{2}x\cos {{x}^{3}}\cos ec\,{{x}^{2}}\]

\[\frac{3}{2}x\cos {{x}^{3}}\cos ec\,{{x}^{2}}\]

\[\frac{3}{2}x\sec {{x}^{3}}\cos ec\,{{x}^{2}}\]

If \[u=f({{x}^{3}}),v=g({{x}^{2}}),f'(x)=\cos x\] and \[g'(x)=\sin x,\

1.	If \[u=f({{x}^{3}}),v=g({{x}^{2}}),f'(x)=\cos x\] and \[g'(x)=\sin x,\] then \[\frac{du}{dv}=\]
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.	\[\frac{3}{2}x\sec {{x}^{3}}\cos ec\,{{x}^{2}}\]
Answer» C. \[\frac{1}{2}x\sec {{x}^{3}}\sin \,{{x}^{2}}\]

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If \[u=f({{x}^{3}}),v=g({{x}^{2}}),f'(x)=\cos x\] and \[g'(x)=\sin x,\] then \[\frac{du}{dv}=\]

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