Question

If $y=\sqrt{x}+\frac{1}{\sqrt{x}},$ then $\frac{dy}{dx}$ at x=1 is

• A. 1
  
• B. $\frac{1}{2}$
  
• C. $\frac{1}{\sqrt{2}}$
  

Answer 1

$y=\sqrt{x}+\frac{1}{\sqrt{x}}$
$=x^{\frac{1}{2}}+x^{-\frac{1}{2}}$
Differentiating both sides with respect to x, we get
$\frac{dy}{dx}=\frac{d}{dx}(x^{\frac{1}{2}}+x^{-\frac{1}{2}})$
$=\frac{d}{dx}\left(x^{\frac{1}{2}}\right)+\frac{d}{dx}\left(x^{-\frac{1}{2}}\right)$
$=\frac{1}{2}{x}^{\frac{1}{2}-1}+(-\frac{1}{2})x^{\frac{1}{2}-1}$
$(y=x^n\Rightarrow \frac{dy}{dx}=n{x}^{n-1})$
$=\frac{1}{2}{x}^{-\frac{1}{2}}-\frac{1}{2}{x}^{-\frac{3}{2}}$
Putting x=1, we get
${\left(\frac{dy}{dx}\right)}_{x=1}=\frac{1}{2}\times 1-\frac{1}{2}\times 1=0$
Thus, $\frac{dy}{dx}$ at $x=1$ is 0.
Hence, the correct answer is option (d).