Question

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evaluate \\( \frac { d } { d x } \arccos ( - 2 x ) \\) at \\( x = \frac { 1 } { 4 } \\).
use an exact expression.

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:

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