For implicit differentiation we should remember that:
#d/dx{f(x)}=f'(x)# and #d/dy{f(y)}=dy/dxf'{y}#
We can differentiate functions of #x# normally but when differentiating functions of #y# we must remember to apply the chain rule and pull out #dy/dx#.
In this case we just differentiate term by term along the equation:
#5y^2=e^(xy)-3x#
The derivative of the first term #5y^2# is simply #10ydy/dx#
For the second term: #e^(xy)# we need to find the derivative of the exponent and then that multiply by the original function get the derivative.
So, to differentiate #xy# we use the product rule to get:
#y+xdy/dx#
Now multiply that by the original expression to get:
#(y+xdy/dx)e^(xy)#
And finally the last term: #3x# just differentiates to #3#.
So differentiating the equation gives us:
#10ydy/dx = (y+xdy/dx)e^(xy) + 3#
Now we do a bit of re arranging to det #dy/dx# on one side of the equation:
#10ydy/dx = ye^(yx)+xdy/dxe^(xy) + 3#
#10ydy/dx-xdy/dxe^(yx)=ye^(yx)+3#
#dy/dx(10y-xe^(yx))=ye^(yx)+3#
#->dy/dx = (ye^(yx)+3)/(10y-xe^(xy))#
