Some curves are described by equations of the form \(F(x,y)=0\). For example, \(x^2+y^2=1\) is the equation of circle of radius 1. For the upper semi-circle we can solved it for \(y\) and write \(y=\sqrt{1-x^2}\) and for the lower semi-circle \(y=-\sqrt{1-x^2}\). There is no function \(y=f(x)\) near \((\pm 1,0)\) that  satisfies \(F(x,f(x))=0\).

Let \(F(x,y)=0\). Suppose \(F(x_0,y_0)=0\) and \(F\) has continuous first partial derivatives and so is differentiable. There are two questions that we try to  answer:

  1. Can we solve for \(y\) as a function of \(x\) near \((x_0,y_0)\)? In other words, can we find a function \(y=f(x)\) defined on some interval \(I=(x_0-h,x_0+h)\) (for \(h>0\)) such that \(F(x,f(x))=0\)?
  2. If such a function exists, what is \(f'(x_0)=\frac{dy}{dx}\Big|_{x=x_0}?\)

Suppose such a function exists. To find \(f'(x_0)\)

Method (a): We use the chain rule:
\[\frac{dF}{dx}=\frac{\partial F}{\partial x}\underbrace{\frac{dx}{dx}}_{=1}+\frac{\partial F}{\partial
y}\underbrace{\frac{dy}{dx}}_{=f'(x)}=0.\] so that

\[\bbox[#F2F2F2,5px,border:2px solid black]{f'(x_0)=\frac{dy}{dx}\Big|_{x=x_0}=-\frac{\dfrac{\partial F}{\partial x}(x_0,y_0)}{\dfrac{\partial
F}{\partial y}(x_0,y_0)},\quad\text{provided}\quad \frac{\partial F}{\partial y}(x_0,y_0)\neq 0.}\]

Method (b): Because \(F\) is constant and therefore its the total differential is zero.
\[dF=\frac{\partial F}{\partial x} dx+\frac{\partial F}{\partial y} dy=0.\] If we divide the above equation by \(dx\) and assume \(F_y(x_0,y_0)\neq 0\), we obtain the same result as method (a).

We assumed that a function \(y=f(x)\) existed and then showed the condition \(\frac{\partial F}{\partial y}(x_0,y_0)\neq 0\) was required for calculation of \(f'(x_0)\). In fact, it can be proved this condition, \(F_y(x_0,y_0)\neq 0\), is sufficient for the existence of \(y=f(x)\) with the aforementioned conditions. The condition \(F_y(x_0,y_0)\neq 0\) means that the tangent line to the level curve \(F(x,y)=0\) is not vertical, and therefore, a part of the level curve — close enough to the point \((x_0,y_0)\)— can be the graph of the function \(y=f(x)\). When \(F_y(x_0,y_0)=0\), you may not be able to find such a function. For example in Fig. 1, if we just keep the shaded disk around \((x_0,y_0)\) and remove the rest of the level curve, what we get can be the graph of a function, because any vertical line now does not intersect this part of the curve more than once. However, at \(A\) or \(B\), where the tangent line is vertical, we cannot find a disk around them (to keep the curve and remove the rest) where a vertical line does not intersect the level curve twice. Therefore the level curve near \(A\) or \(B\) cannot be the graph of a function.

Figure 1.

Noting \(F_x\) and \(F_y\) are both functions of \(x\) and \(y\), higher derivatives of \(y\) with respect to \(x\) can be found by successive differentiation with respect to \(x\) provided higher partial derivatives of \(F(x,y)\) exist:
\[\frac{\partial F}{\partial x}+\frac{\partial F}{\partial y} y’=0 \Rightarrow \frac{\partial^2 F}{\partial
x^2}+2\frac{\partial^2 F}{\partial x \partial y}y’+\frac{\partial^2 F}{\partial y^2} y’^2+\frac{\partial
F}{\partial y} y”=0\] \[\Rightarrow \frac{\partial^3 F}{\partial x^3}+3\frac{\partial^3 F}{\partial x^2 \partial y} y’+3
\frac{\partial^3 F}{\partial x \partial y^2}y’^2+\frac{\partial^3 F}{\partial y^3}y’^3+3\frac{\partial^2
F}{\partial y^2}y’ y”+3\frac{\partial^2 F}{\partial x \partial y}y”+\frac{\partial
F}{\partial y} y”’=0\]

Example 1
Given \(y^3 – x^3 – 3 y^2 – x^2 + 6=0\), find \(dy/dx\).

Let \(F(x,y)=y^3 – x^3 – 3 y^2 – x^2 + 6\). Thus:
\[F_x(x,y)=3x^2-2x,\quad F_y(x,y)=3y^2-6y,\] and it follows that
\[\frac{dy}{dx}=-\frac{F_x}{F_y}=-\frac{3x^2-2x}{3y^2-6y}.\] We cannot use the above formula when \(3y^2-6y=0\).
\[3y^2-6y=3y(y-2)=0\Rightarrow y=0\ \text{or}\ y=2.\] As you can see in Fig. 2, when \(y=0\) and \(y=2\), the tangent line to the curve is vertical.

Figure 2.

Now suppose the equation \(F(x,y,z)=0\) is given, where \(F\) has continuous partial derivatives. If \(F(x_0,y_0,z_0)=0\) and \(\partial F/\partial z (x_0,y_0,z_0)\neq 0\), \(z\) near \((x_0,y_0,z_0)\) can be written as a function of \(x\) and \(y\), namely \(z=f(x,y)\). In other words, the level surface \(F(x,y,z)=0\) can be locally the graph of a function \(z=f(x,y)\). To find the partial derivatives of \(f\), we differentiate the equation \(F(x,y,z)=0\) with respect to \(x\) and \(y\):
\[\frac{\partial F}{\partial x}+\frac{\partial F}{\partial z}\frac{\partial z}{\partial x}=0,\quad
\frac{\partial F}{\partial y}+\frac{\partial F}{\partial z}\frac{\partial z}{\partial y}=0.\] Therefore:

\[\frac{\partial z}{\partial x}(x_0,y_0)=\frac{\partial f}{\partial x}(x_0,y_0)=-\dfrac{\frac{\partial
F}{\partial x}(x_0,y_0,z_0)}{\frac{\partial F}{\partial z}(x_0,y_0,z_0)}\] and
\[\frac{\partial z}{\partial y}(x_0,y_0)=\frac{\partial f}{\partial y}(x_0,y_0)=-\dfrac{\frac{\partial
F}{\partial y}(x_0,y_0,z_0)}{\frac{\partial F}{\partial z}(x_0,y_0,z_0)}.\]

Example 2
Given \(z^2-x^2-y^2=3\), find \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\).

Let \(F(x,y,z)=z^2-x^2-y^2-3\). Then \[\frac{\partial F}{\partial x}=-2x, \quad \frac{\partial F}{\partial y}=-2y,\quad \frac{\partial F}{\partial z}=2z.\] Therefore:
\[\bbox[#F2F2F2,5px,border:2px solid black]{\frac{\partial z}{\partial x}=-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial
z}}=-\frac{-2x}{2z}=\frac{x}{z},\quad \frac{\partial z}{\partial y}=-\frac{\frac{\partial F}{\partial
y}}{\frac{\partial F}{\partial z}}=-\frac{-2y}{2z}=\frac{y}{z}.}\] Note that \(\frac{\partial F}{\partial z}=2z\neq 0\) and the above expressions are valid everywhere because for any point on the level surface, \(|z|\ge 3\) (note \(z^2=3+x^2+y^2\geq 3\)).

In general we have the following theorem.

Theorem 1. (Implicit Function Theorem) Suppose \(F(x_1,\cdots,x_n,z)=0\) and \(F\) is of class \(C^1\) (i.e., has continuous first partial derivatives). Assume:
\[F(\mathbf{x}_0,z_0)=0\quad \text{and} \quad \frac{\partial F}{\partial z}(\mathbf{x}_0,z_0)\neq 0,\] where \(\mathbf{x}_0\in\mathbb{R}^n\) and \(z_0\in\mathbb{R}\). Then there is a neighborhood \(U\) of \(\mathbf{x}_0\) in \(\mathbb{R}^n\), a neighborhood \(V\) of \(z_0\) in \(\mathbb{R}\), and a function \(f:U\subseteq \mathbb{R}^n\to V\) of class \(C^1\) such that if \(\mathbf{x}=(x_1,\cdots,x_n)\in U\) and \(z\in V\) satisfy \(F(\mathbf{x},z)=0\), then \(z=f(\mathbf{x})\). The partial derivatives of \(f\) are given by:
\[\frac{\partial f}{\partial x_i}(\mathbf{x}_0)=-\dfrac{\dfrac{\partial F}{\partial
x_i}(\mathbf{x}_0,z_0)}{\dfrac{\partial F}{\partial z}(\mathbf{x}_0,z_0)}.\]

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