\[\frac{d}{dx}\left[ {{\tan }^{-1}}\left( \frac{a-x}{1+ax} \right) \right]=\] [Karnataka CET 2001; Pb. CET 2001]

\[\frac{1}{1+{{a}^{2}}}-\frac{1}{1+{{x}^{2}}}\]

\[\frac{1}{1+{{\left( \frac{a-x}{1+ax} \right)}^{2}}}\]

\[\frac{-1}{\sqrt{1-{{\left( \frac{a-x}{1+ax} \right)}^{2}}}}\]

\[\frac{d}{dx}\left[ {{\tan }^{-1}}\left( \frac{a-x}{1+ax} \right) \ri

1.	\[\frac{d}{dx}\left[ {{\tan }^{-1}}\left( \frac{a-x}{1+ax} \right) \right]=\] [Karnataka CET 2001; Pb. CET 2001]
A.	\[-\frac{1}{1+{{x}^{2}}}\]
B.	\[\frac{1}{1+{{a}^{2}}}-\frac{1}{1+{{x}^{2}}}\]
C.	\[\frac{1}{1+{{\left( \frac{a-x}{1+ax} \right)}^{2}}}\]
D.	\[\frac{-1}{\sqrt{1-{{\left( \frac{a-x}{1+ax} \right)}^{2}}}}\]
Answer» B. \[\frac{1}{1+{{a}^{2}}}-\frac{1}{1+{{x}^{2}}}\]

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\[\frac{d}{dx}\left[ {{\tan }^{-1}}\left( \frac{a-x}{1+ax} \right) \right]=\] [Karnataka CET 2001; Pb. CET 2001]

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