Differentiation Question 125
Question: Derivative of $ {{( \sqrt{x}+\frac{1}{\sqrt{x}} )}^{2}} $ is
Options:
A) $ \frac{1}{x^{2}} $
B) $ 1-\frac{1}{x^{2}} $
C) 1
D) $ 1+\frac{1}{x^{2}} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] we have, $ \frac{d}{dx}{ {{( \sqrt{x}+\frac{1}{\sqrt{x}} )}^{2}} } $
$ =\frac{d}{dx}{ x+\frac{1}{x}+2 } $
$ =\frac{d}{dx}(x)+\frac{d}{dx}({x^{-1}})+\frac{d}{dx}(2)=1+(-1){x^{-2}}+0 $
$ =1-\frac{1}{x^{2}} $
BETA
