Let's say we have:
$$ x^2 +y^2 = 25 $$
definition of differentiation: $$\lim_{x\to h} = \frac{f(x+h)-f(x)}{h}=\frac{dy}{dx}$$
when we do implicit differentiation we have:
$$ df(x^2 +y^2) =df(25) $$?
we are differentiating with respect to what? I need the intuition for implicit differentiation of what is going on with the geometry when we perform implicit differentiation.
is there any connection with partial differentiation?
If you have a multivariable function you don't have, e.g. $f(x^2+y^2)$, but $f(x,y)$. $f(x^2+y^2)$ actually means $f(g(x,y))$ where $g$ is a multi-valued function and $f$ is a one variable function having the range of $g$ as its domain. (This effectively makes $f$ a multi-variable function, too, but I seek to avoid the multivariable chain rule here, so lets ignore that for right now!).
Consider a circle. $x^2+y^2=R^2$.
Then $y=\pm \sqrt{R^2-x^2}$
$(x+\Delta x)^2+(y+\Delta y)^2=x^2+y^2+2x\Delta x+2y\Delta y=R^2+2x\Delta x + 2y\Delta y$
So
$[(x+\Delta x)+(y+\Delta y)^2]-(x^2+y^2)=2x\Delta x+ 2y \Delta y$
But we are constrained to the circle. $(x+\Delta x, y+\Delta y)$ is also a point on the circle, so
$(x+\Delta x)^2+(y+\Delta y)^2=R^2 $ as well.
But this can only happen if $2x\Delta x + 2y\Delta y=0\implies \frac{\Delta y}{\Delta x}=\frac{-x}{y}$
Consider implicit differentiation:
$x^2+y^2=25$
$2x+2y\cdot y'=0$
$y'=\frac{-x}{y}$
Regular Differentiation:
$y=\pm \sqrt{R^2-x^2}$
$y'=\frac{-2x}{2\sqrt{R^2-x^2}}=\frac{-x}{\sqrt{R^2-x^2}}=\frac{-x}{y}$
Or basic geometry. A line tangent to a circle is perpendicular a radius joining the center of the circle to that point.
Line $y=mx$ passing through the origin will intersect a circle at $x^2+y^2=R^2$ or $x^2(1+m^2)=R^2\implies x^2=\frac{R^2}{1+m^2}$
Then $y^2=R^2-x^2\implies y^2= \frac{m^2R^2}{1+m^2}$
All this implies $m=y/x$ is the slope of the radius between the center and some point on the circle. The slope perpendicular to a line of slope $m$ is $-1/m$ which in this case is $-x/y$
These principles can be utilized in three dimensions where $z=x^2+y^2$ and the partial derivatives describe properties of related geometric objects as well. $z=x^2+y^2$ is a paraboloid. $z=25$ is a plane cutting it perpendicular to its axis of symmetry. So now you can also use tangent planes to discuss geometric properties of the paraboloid. Since you have two independent variables, you now have two slopes worry about at each point and if $z$ is a constant, those slopes tell you something about tangent lines within that plane.
For more rigor, try here : Implicit Function Theorem
What's going on here can be generalized.
Whether $x^2+y^2=25$ or $x^3\sin(xy)+e^{-2xy}=-45x$ is your function, you can set it up in the form $f(x,y)=0$.
So $x^2+y^2=25\implies x^2+y^2-25=0$
If you have $f(x,y)=0$, the total derivative is $df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy$
But since $f$ is a constant, i.e. 0, $df \equiv 0$.
$0\equiv \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y} dy \implies \frac{dy}{dx}=\frac{-\partial f /\partial x}{\partial f/\partial y}$
If $f(x,y)=x^2+y^2-25$, then $\partial f/\partial x = 2x$ and $\partial f /\partial y =2y$.
So $\frac{dy}{dx} =\frac{-x}{y}$
So yes, very close connection with partial differentiation. Good call on that.
This isn't developed rigorously and misses some important details like how multivariable limits work, but is hopefully good enough for some intuition