MCQOPTIONS
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| 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. \[\frac{x}{({{a}^{2}}+{{b}^{2}})}\left[ \frac{1}{\sqrt{{{x}^{2}}+{{a}^{2}}}}-\frac{2}{\sqrt{{{x}^{2}}+{{b}^{2}}}} \right]\] | |