The Implicit Function Theorem Examples 1

# The Implicit Function Theorem Examples 1

Be sure to review The Implicit Function Theorem page before looking at the examples given below.

## Example 1

Find $\left ( \frac{\partial x}{\partial y} \right )_u$ if $\left\{\begin{matrix} xyuv = 1\\ x + y + u + v = 0 \end{matrix}\right.$.

Let $F(x, y, u, v) = xyuv - 1 = 0$ and $G(x, y, u, v) = x + y + u + v = 0$. The notation $\left ( \frac{\partial x}{\partial y} \right )_u$ implies that the variables $x$ and $v$ are dependent, while the variables $y$ and $u$ are independent.

(1)
\begin{align} \quad \left ( \frac{\partial x}{\partial y} \right )_u = -\frac{\frac{\partial (F, G)}{\partial(y, v)}}{\frac{\partial (F, G)}{(x, v)}} \end{align}

Let's compute these Jacobians. We have that the Jacobian on the numerator is:

(2)
\begin{align} \quad \frac{\partial (F, G)}{\partial (y, v)} = \begin{vmatrix} \frac{\partial F}{\partial y} & \frac{\partial F}{\partial v}\\ \frac{\partial G}{\partial y} & \frac{\partial G}{\partial v} \end{vmatrix} = \begin{vmatrix} xuv & xyu\\ 1 & 1 \end{vmatrix} = xuv - xyu \end{align}

The Jacobian in the denominator is:

(3)
\begin{align} \quad \frac{\partial (F, G)}{\partial (x, v)} = \begin{vmatrix} \frac{\partial F}{\partial x} & \frac{\partial F}{\partial v}\\ \frac{\partial G}{\partial x} & \frac{\partial G}{\partial v} \end{vmatrix} = \begin{vmatrix} yuv & xyu \\ 1 & 1 \end{vmatrix} = yuv - xyu \end{align}

Therefore we have that:

(4)
\begin{align} \quad \left ( \frac{\partial x}{\partial y} \right)_u = -\frac{xuv - xyu}{yuv - xyu} = -\frac{xu(v - y)}{yu(v - x)} \frac{x(v - y)}{y(v - x)} \end{align}

## Example 2

Find $\left ( \frac{\partial x}{\partial y} \right )_z$ if $\left\{\begin{matrix} x^2 + y^2 + z^2 + w^2 = 1\\ x + 2y + 3z + 4w = 2 \end{matrix}\right.$.

Let $F(x, y, z, w) = x ^2 + y^2 + z^2 + w^2 - 1 = 0$ and let $G(x, y, z, w) = x + 2y + 3z + 4w - 2 = 0$. The notation $\left ( \frac{\partial x}{\partial y} \right )_z$ implies that $x$ and $w$ are dependent variables and $y$ and $z$ are independent variables. We have that then:

(5)
\begin{align} \quad \left ( \frac{\partial x}{\partial y} \right )_z = - \frac{\frac{\partial (F, G)}{\partial (y, w)}}{\frac{\partial (F, G)}{\partial (x, w)}} \end{align}

Let's compute these Jacobians. The Jacobian in the numerator is given by:

(6)
\begin{align} \quad \frac{\partial (F, G)}{\partial (y, w)} = \begin{vmatrix} \frac{\partial F}{\partial y} & \frac{\partial F}{\partial w}\\ \frac{\partial G}{\partial y} & \frac{\partial G}{\partial w}\\ \end{vmatrix} = \begin{vmatrix} 2y & 2w\\ 2 & 4\\ \end{vmatrix} = 8y - 4w \end{align}

The Jacobian in the denominator is given by:

(7)
\begin{align} \quad \frac{\partial (F, G)}{\partial (x, w)} =\begin{vmatrix} \frac{\partial F}{\partial x} & \frac{\partial F}{\partial w}\\ \frac{\partial G}{\partial x} & \frac{\partial G}{\partial w}\\ \end{vmatrix} = \begin{vmatrix} 2x & 2w\\ 1 & 4\\ \end{vmatrix} = 8x - 2w \end{align}

Therefore we have that:

(8)
\begin{align} \quad \left ( \frac{\partial x}{\partial y} \right )_z = -\frac{8y - 4w}{8x - 2w} = \frac{4w - 8y}{8x - 2w} = \frac{2w - 4y}{4x - w} \end{align}