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

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Answer 1 · 2016-12-12T16:11:23

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

We use,

$sin^2theta+cos^2theta=1$

Let $y=3arccos(x/2)$

Rearranging

$y/3=arccos(x/2)$

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

$cos^2(y/3)+sin^2(y/3)=1$

$sin^2(y/3)=1-cos^2(y/3)=1-x^2/4$

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

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)=3 arccos (x/2)?