Question

If $f\left(x\right)=\sqrt{x+3-4\sqrt{x-1}}$ then $f^{\prime }\left(x\right)$ at $x=2$ is

• A. 0
• B. $-1$
• C. $-\sqrt{3}$
• D. -4

Answer 1

The correct option is B $-1$

Given $f\left(x\right)=\sqrt{x+3-4\sqrt{x-1}}$

$⟹f^{\prime }\left(x\right)=\frac{d}{dx}\sqrt{x+3-4\sqrt{x-1}}$

$⟹f^{\prime }\left(x\right)=\frac{1}{2\sqrt{x+3-4\sqrt{x-1}}}\times \frac{d}{dx}(x+3-4\sqrt{x-1})$

$⟹f^{\prime }\left(x\right)=\frac{1}{\sqrt{x+3-4\sqrt{x-1}}}\times (1-4\frac{d}{dx}\sqrt{x-1})$

$⟹f^{\prime }\left(x\right)=\frac{1}{\sqrt{x+3-4\sqrt{x-1}}}\times (1-\frac{4}{2\sqrt{x-1}})$

$⟹f^{\prime }\left(2\right)=\frac{1}{\sqrt{2+3-4\sqrt{2-1}}}\times (1-\frac{4}{2\sqrt{2-1}})$

$⟹f^{\prime }\left(2\right)=\frac{1}{\sqrt{2+3-4\sqrt{2-1}}}\times (1-\frac{4}{2\sqrt{2-1}})$

$⟹f^{\prime }\left(2\right)=\frac{1}{\sqrt{5-4}}\times (1-\frac{4}{2})$

$⟹f^{\prime }\left(x\right)=1\times -1=-1$