1.13 Implicit Differentiation
Suppose an equation \(F(x,y) =0\) defines \(y\) implicitly as a differentiable function of \(x\), so that \(F(x,y(x)) = 0\). Differentiating with respect to \(x\) by the chain rule, \[\frac {\partial F}{\partial x}\frac {dx}{dx} +\frac {\partial F}{\partial y} \frac {dy}{dx} = 0,\qquad \text {so}\qquad \frac {dy}{dx}=-\frac {F_x}{F_y}\] wherever \(F_y\neq 0\). The derivative is found without solving for \(y\).
Implicit Function Theorem for One Equation
Theorem 1.13.1. Let \(F\) be defined on a disc containing \((a,b)\), with \(F_x\) and \(F_y\) continuous on the disc, and suppose \(F(a,b) =0\) and \(F_y(a,b) \neq 0\). Then near \((a,b)\) the equation \(F(x,y) = 0\) defines \(y\) as a differentiable function of \(x\), with \(y(a) = b\), and \[\frac {dy}{dx}=-\frac {F_x}{F_y} .\]
The condition \(F_y\neq 0\) is what fails at the points \((\pm 2,0)\) of the ellipse in Example 1.11.1: there \(F = x^2 + 4y^2 - 4\) has \(F_y = 8y = 0\), and the curve has a vertical tangent.
Solution. Let \(F(x,y) = x^3 + y^3 - 6xy\). Then \(F_x = 3x^2 - 6y\) and \(F_y = 3y^2 - 6x\), so \[\frac {dy}{dx}=-\frac {F_x}{F_y} = -\frac {x^2 - 2y}{y^2 -2x} = \frac {2y - x^2}{y^2 - 2x} .\]
One Equation in Three Variables
If \(F(x,y,z) = 0\) defines \(z\) implicitly as a function of \(x\) and \(y\), then differentiating with respect to \(x\), with \(y\) held fixed, \[\frac {\partial F}{\partial x}\frac {\partial x}{\partial x} + \frac {\partial F}{\partial y}\frac {\partial y}{\partial x} + \frac {\partial F}{\partial z} \frac {\partial z}{\partial x} = 0 .\] Here \(\frac {\partial x}{\partial x} = 1\) and \(\frac {\partial y}{\partial x} = 0\), so \(F_x + F_z\, \frac {\partial z}{\partial x} =0\). In the same way with respect to \(y\), and so \[\frac {\partial z}{\partial x} = -\frac {F_x}{F_z},\qquad \frac {\partial z}{\partial y} = -\frac {F_y}{F_z},\qquad F_z\neq 0 .\] The same result comes from differentials. The total differential of \(F\) is \(F_x\,dx + F_y\,dy + F_z\,dz\), and on the surface \(F = 0\) it vanishes, so \[dz = -\frac {F_x}{F_z}\,dx - \frac {F_y}{F_z}\,dy .\] Comparing with \(dz = z_x\,dx + z_y\,dy\) gives the same two formulas.
Example 1.13.3. Find \(\frac {\partial z}{\partial x}\) and \(\frac {\partial z}{\partial y}\) if \(x^3 + y^3 + 6xyz + z^3 = 1\).
Solution. Let \(F(x,y,z) = x^3 + y^3 + 6xyz + z^3 - 1\). Then \[\frac {\partial z}{\partial x} = -\frac {F_x}{F_z} = -\frac {3x^2 + 6yz}{6xy + 3z^2},\qquad \frac {\partial z}{\partial y} = -\frac {F_y}{F_z} = -\frac {3y^2 + 6xz}{6xy + 3z^2},\] wherever \(6xy + 3z^2\neq 0\).
Two Equations in Four Variables
Suppose now that two equations \[F(x,y,z,w) = 0,\qquad G(x,y,z,w) = 0\] define \(z = g(x,y)\) and \(w = f(x,y)\). On the solution set both differentials vanish: \begin {align*} F_z\,dz + F_w\,dw & = -F_x\,dx - F_y\,dy,\\ G_z\,dz + G_w\,dw & = -G_x\,dx - G_y\,dy . \end {align*}
This is a linear system for \(dz\) and \(dw\), and by Cramer’s rule \[dz = \frac { \begin {vmatrix} -F_x\,dx -F_y\,dy & F_w\\ -G_x\,dx -G_y\,dy & G_w \end {vmatrix} }{ \begin {vmatrix} F_z & F_w\\ G_z & G_w \end {vmatrix} },\qquad dw = \frac { \begin {vmatrix} F_z & -F_x\,dx - F_y\,dy\\ G_z & -G_x\,dx - G_y\,dy \end {vmatrix} }{ \begin {vmatrix} F_z & F_w\\ G_z & G_w \end {vmatrix} },\] provided the denominator, the Jacobian \(\frac {\partial (F,G)}{\partial (z,w)}\), is not zero. Splitting each numerator into its \(dx\) and \(dy\) parts and comparing with \(dz = g_x\,dx + g_y\,dy\) and \(dw = f_x\,dx + f_y\,dy\) gives \[\frac {\partial z}{\partial x} = -\frac {\partial (F,G)/\partial (x,w)}{\partial (F,G)/\partial (z,w)},\qquad \frac {\partial z}{\partial y}= -\frac {\partial (F,G)/\partial (y,w)}{\partial (F,G)/\partial (z,w)},\] \[\frac {\partial w}{\partial x} = -\frac {\partial (F,G)/\partial (z,x)}{\partial (F,G)/\partial (z,w)},\qquad \frac {\partial w}{\partial y}= -\frac {\partial (F,G)/\partial (z,y)}{\partial (F,G)/\partial (z,w)} .\] The pattern is easy to remember: the denominator is always the Jacobian with respect to the dependent variables \((z,w)\), and the numerator is the same Jacobian with the variable being differentiated replaced by the one it is differentiated with respect to.
Example 1.13.4. Find \(\frac {\partial z}{\partial x}\) and \(\frac {\partial w}{\partial y}\) if \(F = xz + yw - 1 = 0\) and \(G = xw - yz = 0\).
Solution. Here \(F_x = z\), \(F_y = w\), \(F_z = x\), \(F_w = y\) and \(G_x = w\), \(G_y = -z\), \(G_z = -y\), \(G_w = x\), so \[\frac {\partial (F,G)}{\partial (z,w)} = \begin {vmatrix} x & y\\ -y & x\end {vmatrix} = x^2 + y^2 .\] Then \[\frac {\partial z}{\partial x} = -\frac {1}{x^2+y^2}\begin {vmatrix} z & y\\ w & x\end {vmatrix} = -\frac {xz - yw}{x^2 + y^2},\qquad \frac {\partial w}{\partial y} = -\frac {1}{x^2+y^2}\begin {vmatrix} x & w\\ -y & -z\end {vmatrix} = \frac {xz - yw}{x^2 + y^2} .\] (The equations say \((x - iy)(z + iw) = 1\) in complex notation, so that \(z = \frac {x}{x^2+y^2}\) and \(w = \frac {y}{x^2+y^2}\), which explains the symmetry of the answers.)
Example 1.13.5. Find the tangent line to the curve \(x^3 + y^3 = 6xy\) at the point \((3,3)\), and the points where the tangent is horizontal.
Solution. By Example 1.13.2, \(y' = \frac {2y - x^2}{y^2 - 2x}\), which at \((3,3)\) is \(\frac {6 - 9}{9 - 6} = -1\). The tangent line is \(y - 3 = -(x - 3)\), or \(x + y = 6\).
The tangent is horizontal where \(2y - x^2 = 0\) and \(y^2 - 2x\neq 0\). Substituting \(y = \frac {x^2}{2}\) into the equation of the curve, \(x^3 + \frac {x^6}{8} = 3x^3\), so \(x^3\left (\frac {x^3}{8} - 2\right ) = 0\). The root \(x = 0\) gives the origin, where \(F_x = F_y = 0\) and the theorem does not apply (two branches of the curve cross there). The other root is \(x^3 = 16\), so \(x = 2^{4/3}\approx 2.52\) and \(y = \frac {x^2}{2} = 2^{5/3}\approx 3.17\).
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