#color(blue)("The teaching bit")#
By example: Suppose we had #y^2+3y=x^4#
Let #z=y^2+3y-> (dz)/(dy)= 2y+3#...................(1)
Then #z=x^4->(dz)/(dx)=4x^3#...............................(2)
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Consider equation (1). If we can convert #(dz)/(dy)# to #(dz)/(dx)# then we can directly equate the Eqn(1) to Eqn(2)
We use:#" "(dz)/(cancel(dy))xx(cancel(dy))/(dx)#
But from Eqn(1)
#(dz)/(dy)=2y+3" "# so #" "(dz)/(dy)xx(dy)/(dx)" "=" "(dy)/(dx)(2y+3)#
so we have #(dy)/(dx)(2y+2)=(dz)/(dx)=4x^3#
Thus #(dy)/(dx)=(4x^3)/(2y+2)#
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#color(blue)("Answering the question")#
Differentiating each term independently
As appropriate use #(d)/(dx)(uv)=v(du)/(dx)+u(dv)/(dx)#
As appropriate use #(d)/(dx)(u/v)=(v(du)/(dx)-u(dv)/(dx))/v^2#
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#d/(dx)(x^2)=color(brown)(+2x)#
#d/(dx)(x/y)=color(brown)( ( y-x(dy)/(dx))/y^2)#
#d/(dx)(-xy^2) = -[y^2+x2y(dy)/(dx)]=color(brown)(-y^2-2xy(dy)/(dx)#
#d/(dx)(x)=color(brown)(+1)#
#d/(dx)(3y) =color(brown)( 3(dy)/(dx))#
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#color(brown)("Putting it all together")#
The differentiated equation is:
#2x+(y-x(dy)/(dx))/y^2 -y^2-2xy(dy)/(dx) +1=3(dy)/(dx)#
Collecting like terms
#3(dy)/(dx) + (x(dy)/(dx))/y^2+2xy(dy)/(dx)=2x+cancel(y)/y^(cancel(2))-y^2+1 #
#(3y^2(dy)/(dx)+x(dy)/(dx)+2xy^3(dy)/(dx))/y^2=2x+1/y-y^2+1#
#(dy)/(dx)(3y^2+x+2xy^3)=y^2(2x+1/y-y^2+1)#
#(dy)/(dx)=(2xy^2+y-y^4+y^2 )/(3y^2+x+2xy^3)#
