What is the derivative of #arcsin x - sqrt (1-x^2)#?

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Answer 1 · 2016-06-21T17:34:08

# (1+x)/(\sqrt{1 - x^2}) #

do it term by term as #y = y_1 - y_2# so #y' = y_1' - y_2'#

for #y_1 = arcsin x, \qquad sin y_1 = x#

so #cos y_1 \ y_1' = 1#

#y_1' = 1/ (cos y_1) = 1/(\sqrt{1 - sin^2 y_1}) = 1/(\sqrt{1 - x^2}) #

for #y_2 = (1-x^2)^{1/2}#, it is simply

#y_2' = 1/2 (1-x^2)^{-1/2} (-2x)#
# = -(x)/ (1-x^2)^{1/2} = -(x)/ sqrt(1-x^2) #

#\implies y' = 1/(\sqrt{1 - x^2}) + (x)/ sqrt(1-x^2) #

# = (1+x)/(\sqrt{1 - x^2}) #