Implicit differentiation: differential vs derivative

Question

When I search implicit differentiation for equation $x^2 + y^2 = r^2$ I find results of two versions: one using derivative and the other using differential.

Version1: $\frac{d }{dx}(x^2 + y^2 = r^2) \Leftrightarrow 2x + 2y\frac{dy}{dx} = 0 $
Version2: $d(x^2 + y^2 = r^2) \Leftrightarrow 2xdx + 2ydy = 0 $

Using both methods, I can derive the result: $\frac{dy}{dx} = -\frac{x}{y}$

However, I am confused, could you please provide some answers to:

• Which one (derivative /differential) is the "real" implicit differentiation?
• What are the differences between using these two methods?
• When should differential be used rather than derivative?

Answer 1

I think the geometric set-up is not really clear. In the first case you are viewing $y$ as a function of $x$, so the equation $x^2+y^2=r^2$ defines two semicircles in $\mathbb R^2$, i.e. $y = \pm (r^2-x^2)^{1/2}$. Then it makes sense to write $dy/dx$, and the resulting equation in version one will be an equation in the only variable $x$.

In the second case the dependence of $y$ on $x$ is not clear. In principle they could be independent variables. You would then have a vanishing function of two variables $x,y$ which is $f(x,y) = x^2+y^2-r^2$, whose differential is zero: $df = 2xdx + 2ydy = 0$. Since $dx,dy$ are independent this gives $x=y=0$, meaning that $(x,y)=(0,0)$ is a critical point of $f$. This is clear geometrically as $f$ represents a cone in $\mathbb R^3$ whose vertex is $(0,0,-r^2)$.

Answer 2

Any two-variable relation can be summarized by the equation $$f(x,y)=0$$ In your example this would be $$x^2+y^2-r^2=0$$ The total derivative of $f$ with respect to an arbitrary variable $u$ is $$\frac{\mathrm{d}f}{\mathrm{d}u} = \frac{\partial f}{\partial x} \frac{\mathrm{d} x}{\mathrm{d} u}+ \frac{\partial f}{\partial y} \frac{\mathrm{d} y}{\mathrm{d} u}$$ We can plug in $u=x$ to obtain what you mention. Implicit differentiation is just another way to view total multivariable differentiation.