Question

Find y' if y=sin(x+y)

Answer 1

y=sin(x+y)
y'=cos(x+y)[d/dx(x+y)]
solve

Answer 2

if you think of \(y=f(x)\) then this is
\[f(x)=\sin(x+f(x))\] taking derivatives using the chain rule gives
\[f'(x)=\cos(x+f(x))(1+f'(x))\] more conveniently written as
\[y'=\cos(x+y)(1+y')\] then use algebra to solve for \(y'\)

Answer 3

y = sin ( x + y )

y' = Cos( x + y ) d/dx ( x + y )

y' = cos( x + y ) ( 1 + y' )

y' = cos ( x + y ) + cos ( x + y ) y'

y' - cos ( x + y ) y' = cos ( x + y )

y' [ 1 - cos( x + y ) ] = cos ( x + y )

y' = cos ( x + y ) / 1 - cos ( x + y )