How do you find the derivative of $g(x)=3 arccos (x/2)$ ?

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Answer 1 · 2015-07-08T22:16:03

Assuming you do not remember the derivative of $arccosu$ , which is $-1/sqrt(1-u^2) ((du)/(dx))$ :

$y = 3arccos(x/2)$

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

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

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

Since $sin^2u + cos^2u = 1$ :

$= -3/(2sqrt(1-cos^2(y/3))$

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

Getting it even simpler:

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

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

$= color(blue)(-3/sqrt(4-x^2))$

How do you find the derivative of g(x)=3 arccos (x/2)g(x)=3 arccos (x/2)?