A) \[\frac{3x}{1+{{x}^{3}}}\]
B) \[\frac{3{{x}^{2}}}{1+{{x}^{6}}}\]
C) \[\frac{-6{{x}^{5}}}{{{(1+{{x}^{6}})}^{2}}}\]
D) \[\frac{-6{{x}^{5}}}{1+{{x}^{6}}}\]
E) \[{{\tan }^{-1}}x\]
Correct Answer: B
Solution :
Given, \[\frac{d}{dx}\{f(x)\}=\frac{1}{1+{{x}^{2}}}\] On integrating both sides, we get \[f(x)={{\tan }^{-1}}x\] \[\therefore \] \[\frac{d}{dx}=f{{(x)}^{3}}=\frac{d}{dx}({{\tan }^{-1}}{{x}^{3}})\] \[=\frac{1}{1+{{({{x}^{3}})}^{2}}}.3{{x}^{2}}\] \[=\frac{3{{x}^{2}}}{1+{{x}^{6}}}\]
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