One must use the chain rule to differentiate #sin(xy)#

#(df(g(x,y)))/dx = (d(f(g)))/(dg)(d(g(x,y)))/(dx)#

let #g(x,y) = xy#, then #f(g) = sin(g), (df(g))/(dg) = cos(g),# And we need to use the product for the last part:

#(d(g(x,y)))/(dx) = (d(xy))/dx#

#(d(uv))/dx = (u')(v) + (u)(v')#

let #u = x#, then #v = y, u' = 1# and #v' = dy/dx#

#(d(g(x,y)))/(dx) = (1)(y) + (x)(dy/dx) = y + xdy/dx#

#(dsin(xy))/dx = cos(xy)(y + xdy/dx)#

#(dsin(xy))/dx = ycos(xy) + xcos(xy)dy/dx#

Differentiating the right side is trivial:

#(d(2x + 5))/dx = 2#

Put these back into their respective locations within the equation:

#ycos(xy) + xcos(xy)dy/dx = 2#

Solve for #dy/dx#:

#xcos(xy)dy/dx = 2 - ycos(xy)#

#dy/dx = (2 - ycos(xy))/(xcos(xy))#