you might need: calculator\nevaluate \\( \\frac { d } { d x } \\arccos ( - 2 x ) \\) at \\( x = \\frac { 1 }…

Explanation:

Step1: Apply the chain rule

The derivative of (y = \arccos(u)) is (y^\prime=-\frac{1}{\sqrt{1 - u^{2}}}\cdot u^\prime). Let (u=-2x), then (u^\prime=-2). So (\frac{d}{dx}[\arccos(-2x)]=-\frac{1}{\sqrt{1-(-2x)^{2}}}\cdot(-2)=\frac{2}{\sqrt{1 - 4x^{2}}}).

Step2: Substitute (x = \frac{1}{4})

When (x=\frac{1}{4}), we have (\frac{2}{\sqrt{1-4\times(\frac{1}{4})^{2}}}=\frac{2}{\sqrt{1 - 4\times\frac{1}{16}}}=\frac{2}{\sqrt{1-\frac{1}{4}}}=\frac{2}{\sqrt{\frac{3}{4}}}). Since (\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}), then (\frac{2}{\sqrt{\frac{3}{4}}}=\frac{2}{\frac{\sqrt{3}}{2}}=\frac{4}{\sqrt{3}}=\frac{4\sqrt{3}}{3}).

Answer:

(\frac{4\sqrt{3}}{3})