trig. implicit differentiation problem

[skyblue]

if x + sin(x+y)=1, then dy/dx =

okay so heres what i did so far:

1 + cos x + cos y = 0
cos y = 1 + cos x
i'm confused as to whether it is correct to distribute the sin to (x+y) and i need help. i know the answer is - sec(x+y)-1.

 

[pka]

\(\displaystyle \begin{array}{l}
x + \sin (x + y) = 1 \\
1 + \cos (x + y)\left[ {1 + y'} \right] = 0 \\
\end{array}\)

 

[Johnwill]

so you need to use the chain rule?

 

[skeeter]

yep.