A) \[\frac{1}{{{x}^{2}}}\]
B) \[1-\frac{1}{{{x}^{2}}}\]
C) 1
D) \[1+\frac{1}{{{x}^{2}}}\]
Correct Answer: B
Solution :
[b] we have, \[\frac{d}{dx}\left\{ {{\left( \sqrt{x}+\frac{1}{\sqrt{x}} \right)}^{2}} \right\}\] \[=\frac{d}{dx}\left\{ x+\frac{1}{x}+2 \right\}\] \[=\frac{d}{dx}(x)+\frac{d}{dx}({{x}^{-1}})+\frac{d}{dx}(2)=1+(-1){{x}^{-2}}+0\] \[=1-\frac{1}{{{x}^{2}}}\]
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