Question

The derivative of $\sqrt{\left(\sqrt{x}+1\right)}$ is

• A. $\frac{1}{\sqrt{x}\sqrt{\left(\sqrt{\mathrm{x}}+1\right)}}$
• B. $\frac{1}{\sqrt{x}\left(\sqrt{\mathrm{x}}+1\right)}$
• C. $\frac{4}{\sqrt{x}\sqrt{\left(\sqrt{\mathrm{x}}+1\right)}}$
• D. $\frac{1}{4\sqrt{\left(\mathrm{x}\sqrt{\mathrm{x}}+1\right)}}$

Answer 1

The correct option is D

$\frac{1}{4\sqrt{\left(\mathrm{x}\sqrt{\mathrm{x}}+1\right)}}$

Explanation for the correct option:

Finding the derivative of the given function:

$$\begin{gathered} \text{Let,}\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\sqrt{\mathrm{x}}+1} \\ \frac{\mathrm{d}\left(\mathrm{f}\right(\mathrm{x}\left)\right)}{\mathrm{dx}}=\frac{\mathrm{d}\left(\sqrt{\sqrt{\mathrm{x}}+1}\right)}{\mathrm{d}(\sqrt{\mathrm{x}}+1)}\times \frac{\mathrm{d}(\sqrt{\mathrm{x}}+1)}{\mathrm{dx}}\text{(Applying Chain Rule)} \\ =\frac{\mathrm{d}(\sqrt{\mathrm{x}}+1{)}^{\frac{1}{2}}}{\mathrm{d}(\sqrt{\mathrm{x}}+1)}\times \left[\frac{\mathrm{d}}{\mathrm{dx}}\left(\sqrt{\mathrm{x}}\right)+\frac{\mathrm{d}}{\mathrm{dx}}\left(1\right)\right] \\ =\frac{1}{2}\times (\sqrt{\mathrm{x}}+1{)}^{\frac{1}{2}-1}\times \left[\frac{1}{2}\times (\mathrm{x}{)}^{\frac{1}{2}-1}+0\right] \\ =\frac{1\times (\sqrt{\mathrm{x}}+1{)}^{-1/2}}{2}\times \left[\frac{1}{2}\times {\mathrm{x}}^{-\frac{1}{2}}\right] \\ =\frac{1}{2(\sqrt{\mathrm{x}}+1{)}^{\frac{1}{2}}}\times \frac{1}{2{\mathrm{x}}^{{\displaystyle \frac{1}{2}}}} \\ =\frac{1}{2\sqrt{\sqrt{\mathrm{x}}+1}}\times \frac{{\displaystyle 1}}{{\displaystyle 2\sqrt{\mathrm{x}}}} \\ =\frac{{\displaystyle 1}}{{\displaystyle 4\sqrt{\mathrm{x}}\sqrt{(\sqrt{\mathrm{x}}+1)}}} \\ =\frac{{\displaystyle 1}}{{\displaystyle 4\sqrt{(\mathrm{x}\sqrt{\mathrm{x}}+1)}}} \end{gathered}$$

Therefore, the correct answer is option (D).