How do you differentiate x+siny=0 with respect to x?

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Answer 1 · 2017-03-08T03:59:44

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

$x + sin y = 0$
$sin y = -x$
$y = sin^(-1) (-x)$

let say $u = -x$ ,
$(d u)/(d x) = -1$

therefore,

$y = sin^(-1) u$
$(d y)/(d x) = 1/(sqrt (1-u^2)) * (du)/(dx)$

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

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