How do you find $dx/dy$ given $y=sinx+cosx$ ?

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Answer 1 · 2016-12-17T00:35:13

$dx/dy = 1/(cosx-sinx)$

$y=sinx+cosx$
$:. dy/dx = cosx-sinx$

By the chain rule $dx/dy = 1/(dy/dx)$

$:. dx/dy = 1/(cosx-sinx)$

How do you find dx/dydx/dy given y=sinx+cosxy=sinx+cosx?