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)#