Finding #dy/dx# by implicit differentiation involves solving an equation. There is no equation here, only an expression.
Here is how you differentiate with respect to #x#, assuming that #y# is a function (or functions) of #x#:
To find #d/dx(1/2ycos(xy^2))#, we'll need the product rule, the derivative of the cosine, the chain rule and the product rule some more.
I use the product rule in this order: #(fg)' = f'g+fg'#, so we write:
#d/dx(1/2ycos(xy^2)) = [d/dx(1/2y)][cos(xy^2)]+[1/2y][d/dx(cos(xy^2))]#
Let's separate finding #d/dx(cos(xy^2))#, because it may be challenging.
#d/dx(cosu) = -sinu(du)/dx# and in this case #u# is a product of #x# and the square of (the function of #x# named #y#). Therefore, we get:
#d/dx(cos(xy^2)) = -sin(xy^2) [d/dx(xy^2)] #
# = -sin(xy^2)[1y^2+2xy dy/dx]#
Returning to the main problem:
#d/dx(1/2ycos(xy^2)) = [d/dx(1/2y)][cos(xy^2)]+[1/2y][d/dx(cos(xy^2))]#
# = 1/2 dy/dx cos(xy^2) + 1/2y[-sin(xy^2)(1y^2+2xy dy/dx)] #
# = 1/2 cos(xy^2) dy/dx + 1/2y[-y^2sin(xy^2) - 2xysin(xy^2) dy/dx)] #
# = 1/2 cos(xy^2) dy/dx - 1/2y^3sin(xy^2) - xy^2sin(xy^2) dy/dx) #
More algebraic rewriting is possible, but we cannot solve for anything because there is no equation.
