Question

Find the derivative of $\sqrt{3}x-3\sqrt{x}+\frac{1}{\sqrt{x}}$

• A. $\sqrt{3}-\frac{1}{2\sqrt{x}}(3+\frac{1}{x})$
• B. $\sqrt{3}-\frac{1}{\sqrt{x}}(3-\frac{1}{2x})$
• C. $\sqrt{3}+\frac{1}{2\sqrt{x}}(3+\frac{1}{x})$
• D. $\sqrt{3}+\frac{1}{2\sqrt{x}}(3-\frac{1}{x})$

Answer 1

The correct option is A $\sqrt{3}-\frac{1}{2\sqrt{x}}(3+\frac{1}{x})$

$f\left(x\right)=\sqrt{3}x-3\sqrt{x}+\frac{1}{\sqrt{x}}$

$f^{\prime }\left(x\right)=\sqrt{3}-3\cdot \frac{1}{2}\cdot \frac{1}{\sqrt{x}}+(-\frac{1}{2})(x{)}^{-\frac{3}{2}}$

$=\sqrt{3}-\frac{3}{2\sqrt{x}}-\frac{1}{2x\sqrt{x}}$

$=\sqrt{3}-\frac{1}{2\sqrt{x}}(3+\frac{1}{x})$

$\therefore$ Hence, option (A) is correct.