1.

If \[f(x)=\frac{1}{\sqrt{{{x}^{2}}+{{a}^{2}}}+\sqrt{{{x}^{2}}+{{b}^{2}}}}\], then \[{f}'(x)\] is equal to [Kurukshetra  CEE 1998]

A.          \[\frac{x}{({{a}^{2}}-{{b}^{2}})}\left[ \frac{1}{\sqrt{{{x}^{2}}+{{a}^{2}}}}-\frac{1}{\sqrt{{{x}^{2}}+{{b}^{2}}}} \right]\]
B.            \[\frac{x}{({{a}^{2}}+{{b}^{2}})}\left[ \frac{1}{\sqrt{{{x}^{2}}+{{a}^{2}}}}-\frac{2}{\sqrt{{{x}^{2}}+{{b}^{2}}}} \right]\]
C.            \[\frac{x}{({{a}^{2}}-{{b}^{2}})}\left[ \frac{1}{\sqrt{{{x}^{2}}+{{a}^{2}}}}+\frac{1}{\sqrt{{{x}^{2}}+{{b}^{2}}}} \right]\]
D.            \[({{a}^{2}}+{{b}^{2}})\left[ \frac{1}{\sqrt{{{x}^{2}}+{{a}^{2}}}}-\frac{2}{\sqrt{{{x}^{2}}+{{b}^{2}}}} \right]\]
Answer» B.            \[\frac{x}{({{a}^{2}}+{{b}^{2}})}\left[ \frac{1}{\sqrt{{{x}^{2}}+{{a}^{2}}}}-\frac{2}{\sqrt{{{x}^{2}}+{{b}^{2}}}} \right]\]


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