## Calculus: Early Transcendentals 8th Edition

$y"=\frac{cosxcos^2y+sin^2xsiny}{cos^3y}$
Find y' by taking the derivative of both sides of the equation: $cosy\times y'-sinx=0$ $y'=\frac{sinx}{cosy}$ Find y": $y"=\frac{(cosy)(cosx)-(sinx)(-siny\times y')}{cos^2y}$ Plug y' into y": $y"=\frac{cosxcosy+sinxsiny\times\frac{sinx}{cosy}}{cos^2y}$ Simplify: $y"=\frac{cosxcos^2y+sin^2xsiny}{cos^3y}$