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

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

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

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

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