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) none of these
Correct Answer: A
Solution :
\[y={{\tan }^{-1}}\left( \frac{a-x}{1+ax} \right)\]put \[a=\tan \theta ,\]\[x=\tan \text{o }\!\!|\!\!\text{ }\] \[y={{\tan }^{-1}}\left( \frac{\tan \theta -\tan \text{o }\!\!|\!\!\text{ }}{1+\tan \theta \operatorname{tano}|} \right)\] \[\Rightarrow \] \[y={{\tan }^{-1}}\tan (\theta -o|)\] \[y={{\tan }^{-1}}a-{{\tan }^{-1}}x\Rightarrow \frac{dy}{dx}=-\frac{1}{1+{{x}^{2}}}\]
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