Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.6 Exercises - Page 372: 41

$g'(x)=\frac{-3}{\sqrt{4-x^2}}$

$g(x)=3\arccos(\frac{x}{2})$ $let$ $u=\frac{x}{2} , u=\frac{1}{2}$ $\frac{d}{dx}(\arccos u)= \frac{-u'}{\sqrt{1-u^2}}$ $g'(u)=3\frac{-u'}{\sqrt{1-u^2}}$ $g'(x)=\frac{-3}{2\sqrt{1-\frac{x^2}{4}}}$ $g'(x)=\frac{-3}{\sqrt{4-x^2}}$