Editing Implicit function theorem (section)

== The circle example ==
Let us go back to the example of the [[unit circle]]. In this case ''n'' = ''m'' = 1 and <math>f(x,y) = x^2 + y^2 - 1</math>. The matrix of partial derivatives is just a 1 × 2 matrix, given by
<math display="block">(Df)(a,b) = \begin{bmatrix} \dfrac{\partial f}{\partial x}(a,b) & \dfrac{\partial f}{\partial y}(a,b) \end{bmatrix} = \begin{bmatrix} 2a & 2b \end{bmatrix}</math>

Thus, here, the {{math|''Y''}} in the statement of the theorem is just the number {{math|2''b''}}; the linear map defined by it is invertible [[if and only if]] {{math|''b'' ≠ 0}}. By the implicit function theorem we see that we can locally write the circle in the form {{math|1=''y'' = ''g''(''x'')}} for all points where {{math|''y'' ≠ 0}}. For {{math|(±1, 0)}} we run into trouble, as noted before. The implicit function theorem may still be applied to these two points, by writing {{mvar|x}} as a function of {{mvar|y}}, that is, <math>x = h(y)</math>; now the graph of the function will be <math>\left(h(y), y\right)</math>, since where {{math|1=''b'' = 0}} we have {{math|1=''a'' = 1}}, and the conditions to locally express the function in this form are satisfied.

The implicit derivative of ''y'' with respect to ''x'', and that of ''x'' with respect to ''y'', can be found by [[Differential of a function#Differentials in several variables|totally differentiating]] the implicit function <math>x^2+y^2-1</math> and equating to 0:
<math display="block">2x\, dx+2y\, dy = 0,</math>
giving
<math display="block">\frac{dy}{dx}=-\frac{x}{y}</math>
and
<math display="block">\frac{dx}{dy} = -\frac{y}{x}. </math>