Question #994ee

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Answer 1 · 2018-01-29T01:23:28

$- \frac{sin [x - y]}{1 - sin [x - y]}$ $-sin[x-y]/[1-sin[x-y]]$

Using the chain rule and implicit differentiation,
$\frac{d y}{d x} = - sin [x - y] \frac{d}{d x} [x - y]$ $dy/dx =-sin[x-y]d/dx[x-y]$

$\frac{d y}{d x} = - sin [x - y] [1 - \frac{d y}{d x}]$ $dy/dx=-sin[x-y][1-dy/dx]$

$- \frac{d y}{d x} / [sin [x - y] + \frac{d y}{d x} = 1$ $-dy/dx/[sin[x-y]+dy/dx=1$

$- \frac{d y}{d x} [\frac{1}{sin [x - y]} - 1] = 1$ $-dy/dx[1/sin[x-y]-1]=1$

$\frac{d y}{d x} = - \frac{1}{\frac{1}{sin [x - y]} + 1}$ $dy/dx=-1/[1/[sin[x-y]]+1]$

$\frac{d y}{d x} = - \frac{1}{1 - sin [x - y]} / [sin [x - y]]$ $dy/dx=-1/[1-sin[x-y]]/[sin[x-y]]$

$\frac{d y}{d x} = - \frac{sin [x - y]}{1 - sin [x - y]}$ $dy/dx=-sin[x-y]/[1-sin[x-y]]$