How do you find the derivative of the function: $Sin(Arc Cosx)$ ?

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Answer 1 · 2016-04-19T18:11:38

I would rewrite using trigonometry, then differentiate

$theta = arccosx$ if and only if $0 <= theta <= pi$ and $cos theta = x$

Therefore $sin theta = sqrt(1-cos^2theta) = sqrt(1-x^2)$

So,

$d/dx(sin(arccos(x))) = d/dx(sqrt(1-x^2))$

$= 1/(2sqrt(1-x^2)) d/dx(1-x^2)$

(using $d/dx(sqrtu) = 1/(2sqrtu) d/dx(u)$ )

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

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

How do you find the derivative of the function: Sin(Arc Cosx)Sin(Arc Cosx)?