Question

Mark the correct alternative in each of the following:

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

(a) 1 (b) $\frac{1}{2}$ (c) $\frac{1}{\sqrt{2}}$ (d) 0

Answer 1

$$\begin{gathered} y=\sqrt{x}+\frac{1}{\sqrt{x}} \\ =x^{\frac{1}{2}}+x^{-\frac{1}{2}} \end{gathered}$$

Differentiating both sides with respect to x, we get

$$\begin{gathered} \frac{dy}{dx}=\frac{d}{dx}\left(x^{\frac{1}{2}}+x^{-\frac{1}{2}}\right) \\ =\frac{d}{dx}\left(x^{\frac{1}{2}}\right)+\frac{d}{dx}\left(x^{-\frac{1}{2}}\right) \\ =\frac{1}{2}{x}^{\frac{1}{2}-1}+\left(-\frac{1}{2}\right)x^{-\frac{1}{2}-1}\left(y=x^n\Rightarrow \frac{dy}{dx}=n{x}^{n-1}\right) \\ =\frac{1}{2}{x}^{-\frac{1}{2}}-\frac{1}{2}{x}^{-\frac{3}{2}} \end{gathered}$$

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).