A) 2/3
B) -2/3
C) 3/2
D) -3/23
Correct Answer: C
Solution :
\[y={{\tan }^{-1}}\left( \frac{3{{x}^{2}}-1}{3x-{{x}^{3}}} \right)\] \[=\frac{\pi }{2}-{{\tan }^{-1}}\left( \frac{3x-{{x}^{3}}}{3{{x}^{2}}-1} \right)=\frac{\pi }{2}+3{{\tan }^{-1}}x\] \[\frac{dy}{dx}=\frac{3}{1+{{x}^{2}}}\] \[u={{\sin }^{-1}}\left( \frac{{{x}^{2}}-1}{{{x}^{2}}+1} \right)=\frac{\pi }{2}-{{\cos }^{-1}}\left( \frac{{{x}^{2}}-1}{{{x}^{2}}+1} \right)=-\frac{\pi }{2}+2{{\tan }^{-1}}x\]\[\frac{du}{dx}=\frac{2}{1+{{x}^{2}}}\]\[\frac{dy}{dx}=\frac{dy/du}{du/dx}=\frac{3}{2}\]
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