A) \[\frac{a}{{{a}^{2}}+{{x}^{2}}}\]
B) \[\frac{-a}{{{a}^{2}}+{{x}^{2}}}\]
C) \[\frac{1}{a\sqrt{{{a}^{2}}-{{x}^{2}}}}\]
D) \[\frac{1}{\sqrt{{{a}^{2}}-{{x}^{2}}}}\]
Correct Answer: D
Solution :
\[\frac{d}{dx}{{\tan }^{-1}}\frac{x}{\sqrt{{{a}^{2}}-{{x}^{2}}}}\] Putting \[x=a\sin \theta ,\] we get \[=\frac{d}{dx}\left[ {{\tan }^{-1}}\frac{a\sin \theta }{a\cos \theta } \right]=\frac{d}{dx}\left[ {{\tan }^{-1}}\tan \theta \right]=\frac{d}{dx}[\theta ]\] Substituting value of\[\theta \], so \[|x|\].
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