#y = -3/5x+7/5# gives #y# explicitly as a function of #x#.

#3x+5y=7# gives exactly the same relationship between #x# and #y#, but the function is implicit (hidden) in the equation. To make the function explicit, we solve for #x#

In #x^2+y^2=25#, #y# is not a function of #x#. However, there are two functions implicit in the equation. We can make the functions explicit by solving for #y#.

#y = +- sqrt(25-x^2)# is equivalent to the equation above and it has 2 functions that are not too difficult to make explicit:

#y=sqrt(25-x^2)# gives #y# as a function of #x# and

#y=-sqrt(25-x^2)# gives #y# as a different function of #x#.

We can differentiate either the implicit or explicit presentations.

Differentiating implicitly (leaving the functions implicit) we get

#2x+2y dy/dx = 0# #" "# so #" "# #dy/dx = -x/y#

The #y# in the formula for the derivative is the price we pay for not making the function explicit. It replaces the explicit form of the function, whatever that may be.

For #y=sqrt(25-x^2)#, we get #dy/dx = - x/sqrt(25-x^2)# (use the power and chain rule), and

for #y= - sqrt(25-x^2)#, we get #dy/dx = x/sqrt(25-x^2)#.

The equation #y^5+4x^2y^2-3y+7x=28 # cannot be solved algebraically for #y#, (or anyway, some 5th degree equations cannot be solved) but there are several functions of #x# implicit in the equation. You can see them in the graph of the equation (shown below).

graph{y^5+4x^2y^2-3y+7x=28 [-7.14, 6.91, -4.66, 2.36]}

We can cut the graph into pieces, each of which is the graph of some function of #x# on some domain.

Implicit differentiation allow us to find the derivative(s) of #y# with respect to #x# **without** making the function(s) explicit. Doing that, we can find the slope of the line tangent to the graph at the point #(1,2)#.