Begin by taking the derivative with respect to both sides:
#d/dx(4)=d/dx(x+y)^2#
On the left, we have the derivative of a constant - which is just #0#. That breaks the problem down to:
#0=d/dx(x+y)^2#
To evaluate #d/dx(x+y)^2#, we need to use the power rule and the chain rule:
#d/dx(x+y)^2=(x+y)'*2(x+y)^(2-1)#
Note: we multiply by #(x+y)'# because the chain rule tells us we have to multiply the derivative of the whole function (in this case #(x+y)^2# by the inside function (in this case #(x+y)#).
#d/dx(x+y)^2=(x+y)'*2(x+y)#
As for #(x+y)'#, notice that we can use the sum rule to break it into #x'+y'#. #x'# is simply #1#, and because we don't actually know what #y# is, we have to leave #y'# as #dy/dx#:
#d/dx(x+y)^2=(1+dy/dx)(2(x+y))#
Now that we've found our derivative, the problem is:
#0=(1+dy/dx)(2(x+y))#
Doing some algebra to isolate #dy/dx#, we see:
#0=(1+dy/dx)(2x+2y)#
#0=2x+dy/dx2x+dy/dx2y+2y#
#0=x+dy/dxx+dy/dxy+y#
#-x-y=dy/dxx+dy/dxy#
#-x-y=dy/dx(x+y)#
#dy/dx=(-x-y)/(x+y)#
Interestingly, this equals #-1# for all #x# and #y# (except when #x=-y#). Therefore, #dy/dx=-1#. We could have actually figured this out without using any calculus at all! Look at the equation #4=(x+y)^2#. Take the square root of both sides to get #+-2=x+y#. Now subtract #x# from both sides, and we have #y=+-2-x#. Remember these from algebra? The slope of this line is #-1#, and since the derivative is the slope, we could have just said #dy/dx=-1# and avoided all that work.
