What is the derivative of $g (x) = 3 arccos (\frac{x}{2})$ $g(x)=3 arccos (x/2)$ ?

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Answer 1 · 2016-12-12T16:11:23

The answer is $= - \frac{3}{2} \cdot \frac{1}{\sqrt{1 - \frac{x^{2}}{4}}}$ $=-3/2*1/sqrt(1-x^2/4)$

We use,

${sin}^{2} θ + {cos}^{2} θ = 1$ $sin^2theta+cos^2theta=1$

Let $y = 3 arccos (\frac{x}{2})$ $y=3arccos(x/2)$

Rearranging

$\frac{y}{3} = arccos (\frac{x}{2})$ $y/3=arccos(x/2)$

$cos (\frac{y}{3}) = \frac{x}{2}$ $cos(y/3)=x/2$

${cos}^{2} (\frac{y}{3}) + {sin}^{2} (\frac{y}{3}) = 1$ $cos^2(y/3)+sin^2(y/3)=1$

${sin}^{2} (\frac{y}{3}) = 1 - {cos}^{2} (\frac{y}{3}) = 1 - \frac{x^{2}}{4}$ $sin^2(y/3)=1-cos^2(y/3)=1-x^2/4$

$sin (\frac{y}{3}) = \sqrt{1 - \frac{x^{2}}{4}}$ $sin(y/3)=sqrt(1-x^2/4)$

Differentiating, we get

$(cos(y/3))'=(x/2)'$

$-1/3sin(y/3)*dy/dx=1/2$

$dy/dx=-3/2*1/sin(y/3)$

$=-3/2*1/sqrt(1-x^2/4)$

What is the derivative of g(x)=3 arccos (x/2)g(x)=3arccos(x2)g(x)=3 arccos (x/2)?