Find the derivative of this implicit function Let y = sin (x + y), x = π, find y ' The answer in this book is - 1 / 2, how to calculate Let y = sin (x + y) determine the implicit function y = y (x), then dy = (2) How did this "2" come from? Can we do that? Y = sin (π + y) → y = - Sin y → derivation on both sides y '= - y'cosy reduces y' to cosy = - 1, but in this way, y '= cos (x + y) * (1 + y') = cos (π + y) * (1 + y ') = - cosy * (1 + y') is sorted out and y '= - cosy / (1 + cosy) is substituted into cosy = - 1, and the result is y' = 1 / O. what's wrong with this process?

Find the derivative of this implicit function Let y = sin (x + y), x = π, find y ' The answer in this book is - 1 / 2, how to calculate Let y = sin (x + y) determine the implicit function y = y (x), then dy = (2) How did this "2" come from? Can we do that? Y = sin (π + y) → y = - Sin y → derivation on both sides y '= - y'cosy reduces y' to cosy = - 1, but in this way, y '= cos (x + y) * (1 + y') = cos (π + y) * (1 + y ') = - cosy * (1 + y') is sorted out and y '= - cosy / (1 + cosy) is substituted into cosy = - 1, and the result is y' = 1 / O. what's wrong with this process?

y'=cos（x+y)(y'+1)
Y '= cos (x + y) / (1-cos (x + y))
When x = π
y=sin(π+y)=sin（-y）
Here y = 0
Y '= cos (x + y) / (1-cos (x + y))
Y '= cos (π) / (1-cos (π)) = - 1 / 2
X is a variable. You made it a constant like that!