1.13 Implicit Differentiation
Suppose that an equation of the form \(F(x,y) =0\) defines \(y\) implicitly i.e \(F(x,f(x)) =0\) \[ \frac {\partial F}{\partial x}\cdot \frac {dx}{dx} +\frac {\partial F}{\partial y}\cdot \frac {dy}{dx} = 0\]
\[\implies \hspace {0.5cm} \frac {dy}{dx}=\frac {-F_x}{F_y}\]
Implicit Function Theorem
If \(F\) is defined on a disc containing \((a,b)\), where \(F(a,b) =0,\hspace {0.2cm} F_y(a,b) \neq 0\) and \(F_x, F_y\) are continuous on the disc then \(F(x,y) =0\) near \((a,b)\) and \[\frac {dy}{dx}=\frac {-F_x}{F_y}\]
Given \(x^3 + y^3 = 6xy\) find \(y'\)
Let \(F(x,y) = x^3 + y^3 - 6xy = 0\)
\begin {align*} \frac {\partial F}{\partial x} & = 3x^2 - 6y\\\\ \frac {\partial F}{\partial y} & = 3y^2 - 6x \end {align*}
\[\implies \hspace {1cm} \frac {dy}{dx}=-\frac {F_x}{F_y} = \frac {-x^2 + 2y}{y^2 -2x}\]
\[F(x,y,z) = 0\]
\[\frac {\partial F}{\partial x}\cdot \frac {\partial x}{\partial x} + \frac {\partial F}{\partial y}\cdot \frac {\partial y}{\partial x} + \frac {\partial F}{\partial z}\cdot \frac {\partial z}{\partial x} = 0\]
\[F_x + F_z\hspace {0.1cm} \frac {\partial z}{\partial x} =0\]
\[\frac {\partial z}{\partial x} = \frac {-F_x}{F_z}\hspace {0.5cm} , \hspace {0.5cm} F_z\neq 0\]
\[\frac {\partial F}{\partial x}\cdot \frac {\partial x}{\partial y} + \frac {\partial F}{\partial y}\cdot \frac {\partial y}{\partial y} + \frac {\partial F}{\partial z}\cdot \frac {\partial z}{\partial y}=0\]
\[F_y + F_z\hspace {0.1cm}\frac {\partial z}{\partial y} = 0\]
\[\frac {\partial z}{\partial y} = \frac {-F_y}{F_z}\hspace {0.5cm},\hspace {0.5cm} F_z\neq 0\]
Recall: Total differentiation of \(z = f(x,y)\) \[dz = f_x \,dx + f_y\, dy\]
Let \(F(x,y,z) = 0\)
Find \(\dfrac {\partial z}{\partial x}\) and \(\dfrac {\partial z}{\partial y}\) if \[x^3 + y^3 + 6xyz + z^3 = 1\]
Let \(F(x,y,z) = x^3 + y^3 + 6xyz + z^3 - 1 =0\) \begin {align*} \frac {\partial z}{\partial x} = \frac {-F_x}{F_z} & = \frac {-\big (3x^2 + 6yz\big )}{\big (6xy + 3z^2\big )}\\\\ \frac {\partial z}{\partial y} = \frac {-F_y}{F_z} & = \frac {-\big (3y^2 + 6xz\big )}{\big (6xy + 3z^2\big )}\\\\ \end {align*}
\[dz = f_x \,dx + f_y\, dy\]
\[F_x\,dx + F_y\,dy + F_z\,dz = 0\]
\[dz = \frac {-F_x}{F_z}\,dx - \frac {F_y}{dz}\,dy\hspace {0.5cm} F_z \neq 0\]
\[\therefore \hspace {0.5cm} \frac {\partial z}{\partial x} = \frac {-F_x}{F_z}\hspace {0.5cm}, \hspace {0.5cm}\frac {\partial z}{\partial y} =\frac {-F_y}{F_z}\]
Suppose \(F(x,y,z,w) = 0\) and \(G(x,y,z,w) = 0\) where \(w = f(x,y),\hspace {0.2cm} z = g(x,y)\) \begin {align*} F_x\,dx + F_y\,dy + F_z\,dz + F_w\,dw & = 0\\\\ G_x\,dx + G_y\,dy + G_z\,dz + G_w\,dw & = 0\\ \end {align*}
Solving for \(dz\) and \(dw\)
\begin {align*} 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} }\hspace {2cm} d_w = \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} }\\ \end {align*}
\[dz = \frac { -\begin {vmatrix} F_x & F_w\\ G_x & G_w\\ \end {vmatrix} \,dx}{ \begin {vmatrix} F_z & F_w\\ G_z & G_w\\ \end {vmatrix} }-\frac { \begin {vmatrix} F_y & F_w\\ G_y & G_w\\ \end {vmatrix} \,dy }{ \begin {vmatrix} F_z & F_w\\ G_z & G_w\\ \end {vmatrix} } \]
\[dz = g_x \,dx + g_y\,dy\]
\[\therefore \hspace {1cm} \frac {\partial z}{\partial x} = \frac {\dfrac {-\partial (F,G)}{\partial (x,w)}}{\dfrac {\partial (F,G)}{\partial (z,w)}}\hspace {1cm} \frac {\partial z}{\partial y}= \frac {\dfrac {-\partial (F,G)}{\partial (y,w)}}{\dfrac {\partial (F,G)}{\partial (z,w)}}\hspace {1cm} \frac {\partial (F,G)}{\partial (z,w)} \neq 0\]
Find \(\dfrac {\partial w}{\partial x}, \dfrac {\partial w}{\partial y}\)
\[dw = \frac { -\begin {vmatrix} F_z & F_x\\ G_z & G_x\\ \end {vmatrix} dx}{ \begin {vmatrix} F_z & F_w\\ G_z & G_w\\ \end {vmatrix} }-\frac { \begin {vmatrix} F_z & F_y\\ G_z & G_y\\ \end {vmatrix} dy }{ \begin {vmatrix} F_z & F_w\\ G_z & G_w\\ \end {vmatrix} } \]
\[dw = f_x\,dx + f_y \,dy\]
\[\therefore \hspace {1cm} \frac {\partial w}{\partial x} = \frac {\dfrac {-\partial (F,G)}{\partial (z,x)}}{\dfrac {\partial (F,G)}{\partial (z,w)}}\hspace {1cm} \frac {\partial w}{\partial y}= \frac {\dfrac {-\partial (F,G)}{\partial (z,y)}}{\dfrac {\partial (F,G)}{\partial (z,w)}}\hspace {1cm} \frac {\partial (F,G)}{\partial (z,w)} \neq 0\]
Determinants are Jacobians.
Questions on this section
Stuck on something here? Ask below and it stays attached to this topic.