How do you find the derivative of the function: #arccos(x^2)#?

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Answer 1 · 2016-04-26T19:00:59

#theta =arccos(x^2)#

#cos(theta) = x^2#

#-theta'sin(theta) = 2x#

#theta' = -(2x)/sin(theta)#

don't forget that #sin(theta) = sqrt(1-cos^2(theta))# and that #cos(theta) = x^2#

so we have

#theta' = -(2x)/sqrt(1-x^4)#