Question

Find the derivative of $\frac{1}{\sqrt{x}}$.

Answer 1

Apply the formula for derivative

As we know that the derivative of $n^{\mathrm{th}}$ power of a variable $x$ is defined as,

$\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}{\mathit{x}}^{\mathbf{n}}\mathbf{}\mathbf{=}\mathbf{}\mathit{n}{\mathit{x}}^{\mathbf{n}\mathbf{-}\mathbf{1}}$

Then, the derivative of $\frac{1}{\sqrt{x}}$ can be calculated as follows,

$\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{\sqrt{x}}\right)=\frac{d}{dx}\left(\frac{1}{x^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\right)$ $\left[\because \sqrt{x}=x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]$

$\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{\sqrt{x}}\right)=\frac{d}{dx}\left(x^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)$ $\left[\because a^{-m}=\frac{1}{a^m}\right]$

$\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{\sqrt{x}}\right)=\left(-\frac{1}{2}\right)\left(x^{\left(-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.-1\right)}\right)$ $\left[\because \frac{d}{dx}{x}^n=n{x}^{n-1}\right]$

$$\begin{gathered} \Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{\sqrt{x}}\right)=\left(-\frac{1}{2}\right)x^{\left(-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)} \\ \Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{\sqrt{x}}\right)=\left(-\frac{1}{2}\right)\frac{1}{x^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \\ \Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{\sqrt{x}}\right)=-\frac{1}{2}\times \frac{1}{{\left(x^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right)}^3} \end{gathered}$$

$\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{\sqrt{x}}\right)=-\frac{1}{2}\times \frac{1}{{\left(\sqrt{x}\right)}^3}$ $\left[\because x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\sqrt{x}\right]$

$$\begin{gathered} \Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{\sqrt{x}}\right)=-\frac{1}{2}\times \frac{1}{x\sqrt{x}} \\ \Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{\sqrt{x}}\right)=-\frac{1}{2x\sqrt{x}} \end{gathered}$$

Hence, the derivative of $\frac{1}{\sqrt{x}}$ is $-\frac{1}{2x\sqrt{x}}$.