Example 4-3-26 - Maple Help

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Chapter 4: Partial Differentiation

Section 4.3: Chain Rule

Example 4.3.26

If the equation $u = x^{3} + y^{3} + z^{3} + u^{3}$ implicitly defines $u = u (x, y, z)$ and the equation $z = x^{2} + y^{2} + z^{2}$ implicitly defines $z = z (x, y)$ , obtain $u_{x}$ and $u_{y}$ .

Solution

Mathematical Solution

Differentiate each equation with respect to $x$ , keeping in mind the implicitly defined functions in each.

$\frac{\partial}{\partial x} (u = x^{3} + y^{3} + z^{3} + u^{3}) \Rightarrow u_{x} = 3 x^{2} + 3 z^{2} z_{x} + 3 u^{2} u_{x} \Rightarrow u_{x} = \frac{3 x^{2} + 3 z^{2} z_{x}}{1 - 3 u^{2}}$

and

$\frac{\partial}{\partial x} (z = x^{2} + y^{2} + z^{2}) \Rightarrow z_{x} = 2 x + 2 z z_{x} = 0 \Rightarrow z_{x} = \frac{2 x}{1 - 2 z}$

from which it follows that $u_{x} = \frac{3 x^{2} + 3 z^{2} (\frac{2 x}{1 - 2 z})}{1 - 3 u^{2}} = \frac{3 x (2 z^{2} - 2 x z + x)}{(3 u^{2} - 1) (2 z - 1)}$ .

Differentiate each equation with respect to $y$ , keeping in mind the implicitly defined functions in each.

$\frac{\partial}{\partial y} (u = x^{3} + y^{3} + z^{3} + u^{3}) \Rightarrow u_{y} = 3 y^{2} + 3 z^{2} z_{y} + 3 u^{2} u_{y} \Rightarrow u_{y} = \frac{3 y^{2} + 3 z^{2} z_{y}}{1 - 3 u^{2}}$

and

$\frac{\partial}{\partial y} (z = x^{2} + y^{2} + z^{2}) \Rightarrow z_{y} = 2 y + 2 z z_{y} \Rightarrow z_{y} = \frac{2 y}{1 - 2 z}$

from which it follows that $u_{y} = \frac{3 y^{2} + 3 z^{2} (\frac{2 y}{1 - 2 z})}{1 - 3 u^{2}} = \frac{3 y (2 z^{2} - 2 y z + y)}{(3 u^{2} - 1) (2 z - 1)}$ .

Maple Solution - Interactive

Obtain $u_{x}$ from first principles

•	Write the first equation with the appropriate dependencies made explicit.

•	Context Panel: Differentiate≻With Respect To≻ $x$

$u (x, y, z) = x^{3} + y^{3} + z {(x, y)}^{3} + u {(x, y, z)}^{3}$

$u (x, y, z) = x^{3} + y^{3} + {z (x, y)}^{3} + {u (x, y, z)}^{3}$

(1)

$\overset{differentiate w.r.t. x}{\to}$

$\frac{\partial}{\partial x} u (x, y, z) = 3 x^{2} + 3 {z (x, y)}^{2} (\frac{\partial}{\partial x} z (x, y)) + 3 {u (x, y, z)}^{2} (\frac{\partial}{\partial x} u (x, y, z))$

(2)

•	Write the second equation with the appropriate dependencies made explicit.

•	Context Panel: Differentiate≻With Respect To≻ $x$

$z (x, y) = x^{2} + y^{2} + z {(x, y)}^{2}$

$z (x, y) = x^{2} + y^{2} + {z (x, y)}^{2}$

(3)

$\overset{differentiate w.r.t. x}{\to}$

$\frac{\partial}{\partial x} z (x, y) = 2 x + 2 z (x, y) (\frac{\partial}{\partial x} z (x, y))$

(4)

•	Using equation labels, make a sequence of the two equations resulting from differentiation, and press the Enter key.

•	Context Panel: Solve≻Solve for Variables≻ $u_{x}, z_{x}$ Enter as per Figure 4.3.26(a).

Figure 4.3.26(a) Variables dialog

$,$

$\frac{\partial}{\partial x} u (x, y, z) = 3 x^{2} + 3 {z (x, y)}^{2} (\frac{\partial}{\partial x} z (x, y)) + 3 {u (x, y, z)}^{2} (\frac{\partial}{\partial x} u (x, y, z)), \frac{\partial}{\partial x} z (x, y) = 2 x + 2 z (x, y) (\frac{\partial}{\partial x} z (x, y))$

(5)

$\overset{solve (specified)}{\to}$

$\{\frac{\partial}{\partial x} u (x, y, z) = \frac{3 x (2 {z (x, y)}^{2} - 2 z (x, y) x + x)}{(2 z (x, y) - 1) (3 {u (x, y, z)}^{2} - 1)}, \frac{\partial}{\partial x} z (x, y) = - \frac{2 x}{2 z (x, y) - 1}\}$

(6)

Thus, $u_{x} = \frac{3 x (2 z^{2} - 2 x z + x)}{(3 u^{2} - 1) (2 z - 1)}$ .

Obtain $u_{y}$ from first principles

•	Write the first equation with the appropriate dependencies made explicit.

•	Context Panel: Differentiate≻With Respect To≻ $y$

$u (x, y, z) = x^{3} + y^{3} + z {(x, y)}^{3} + u {(x, y, z)}^{3}$

$u (x, y, z) = x^{3} + y^{3} + {z (x, y)}^{3} + {u (x, y, z)}^{3}$

(7)

$\overset{differentiate w.r.t. y}{\to}$

$\frac{\partial}{\partial y} u (x, y, z) = 3 y^{2} + 3 {z (x, y)}^{2} (\frac{\partial}{\partial y} z (x, y)) + 3 {u (x, y, z)}^{2} (\frac{\partial}{\partial y} u (x, y, z))$

(8)

•	Write the second equation with the appropriate dependencies made explicit.

•	Context Panel: Differentiate≻With Respect To≻ $y$

$z (x, y) = x^{2} + y^{2} + z {(x, y)}^{2}$

$z (x, y) = x^{2} + y^{2} + {z (x, y)}^{2}$

(9)

$\overset{differentiate w.r.t. y}{\to}$

$\frac{\partial}{\partial y} z (x, y) = 2 y + 2 z (x, y) (\frac{\partial}{\partial y} z (x, y))$

(10)

•	Using equation labels, make a sequence of the two equations resulting from differentiation, and press the Enter key.

•	Context Panel: Solve≻Solve for Variables≻ $u_{y}, z_{y}$ Enter as per Figure 4.3.26(b).

Figure 4.3.26(b) Variables dialog

$,$

$\frac{\partial}{\partial y} u (x, y, z) = 3 y^{2} + 3 {z (x, y)}^{2} (\frac{\partial}{\partial y} z (x, y)) + 3 {u (x, y, z)}^{2} (\frac{\partial}{\partial y} u (x, y, z)), \frac{\partial}{\partial y} z (x, y) = 2 y + 2 z (x, y) (\frac{\partial}{\partial y} z (x, y))$

(11)

$\overset{solve (specified)}{\to}$

$\{\frac{\partial}{\partial y} u (x, y, z) = \frac{3 y (2 {z (x, y)}^{2} - 2 z (x, y) y + y)}{(2 z (x, y) - 1) (3 {u (x, y, z)}^{2} - 1)}, \frac{\partial}{\partial y} z (x, y) = - \frac{2 y}{2 z (x, y) - 1}\}$

(12)

Thus, $u_{y} = \frac{3 y (2 z^{2} - 2 y z + y)}{(3 u^{2} - 1) (2 z - 1)}$ .

Maple Solution - Coded

Apply the implicitdiff command

$implicitdiff (\{u = x^{3} + y^{3} + z^{3} + u^{3}, z = x^{2} + y^{2} + z^{2}\}, \{u, z\}, \{u, z\}, x)$

$\{D_{1} (u) = \frac{3 x (- 2 x z + 2 z^{2} + x)}{(3 u^{2} - 1) (2 z - 1)}, D_{1} (z) = - \frac{2 x}{2 z - 1}\}$

$implicitdiff (\{u = x^{3} + y^{3} + z^{3} + u^{3}, z = x^{2} + y^{2} + z^{2}\}, \{u, z\}, \{u, z\}, y)$

$\{D_{2} (u) = \frac{3 y (- 2 y z + 2 z^{2} + y)}{(3 u^{2} - 1) (2 z - 1)}, D_{2} (z) = - \frac{2 y}{2 z - 1}\}$

From the first calculation, obtain $u_{x} = \frac{3 x (2 z^{2} - 2 x z + x)}{(3 u^{2} - 1) (2 z - 1)}$ ; and from the second, $u_{y} = \frac{3 y (2 z^{2} - 2 y z + y)}{(3 u^{2} - 1) (2 z - 1)}$ .

The following computation of $u_{x}$ and $u_{y}$ from first principles makes use of some notational simplifications. Unfortunately, Maple can either suppress the arguments on $u (x, y, z)$ or $z (x, y)$ , but not both because $z$ would be a suppressed argument of $u$ . The choice here is to suppress the arguments for $u (x, y, z)$ .

Notational simplifications

•	These commands cause $u$ and $u (x, y, z)$ to be equivalent, and for its derivatives to be written with subscripts.

$interface (typesetting = extended) &colon; Typesetting :- Suppress (u (x, y, z)) &colon; Typesetting :- Settings (userep = true) &colon;$

Obtain $u_{x}$

•	Write the two defining equations with the appropriate dependencies explicitly stated. Press the Enter key.

$q_{1} ≔ u (x, y, z) = x^{3} + y^{3} + z^{3} (x, y) + u^{3} (x, y, z) &semi; q_{2} ≔ z (x, y) = x^{2} + y^{2} + z^{2} (x, y)$

$q_{1} ≔ u = x^{3} + y^{3} + {z (x, y)}^{3} + u^{3}$

•	Differentiate both equations with respect to $x$ and solve for the two derivatives $u_{x}$ and $z_{x}$ .

$q_{3} ≔ solve ([diff (q_{1}, x), diff (q_{2}, x)], [diff (u, x), diff (z (x, y), x)])$

$q_{3} ≔ [[u_{x} = \frac{3 x (2 {z (x, y)}^{2} - 2 z (x, y) x + x)}{(2 z (x, y) - 1) (3 u^{2} - 1)}, \frac{\partial}{\partial x} z (x, y) = - \frac{2 x}{2 z (x, y) - 1}]]$

•	Extract the solution for $u_{x}$ and replace $z (x, y)$ with $z$ .

$eval (op ([1, 1], q_{3}), [z (x, y) = z])$

$u_{x} = \frac{3 x (- 2 x z + 2 z^{2} + x)}{(2 z - 1) (3 u^{2} - 1)}$

Obtain $u_{y}$

•	Differentiate both equations with respect to $y$ and solve for the two derivatives $u_{y}$ and $z_{y}$ .

$q_{4} ≔ solve ([diff (q_{1}, y), diff (q_{2}, y)], [diff (u, y), diff (z (x, y), y)])$

$q_{4} ≔ [[u_{y} = \frac{3 y (2 {z (x, y)}^{2} - 2 z (x, y) y + y)}{(2 z (x, y) - 1) (3 u^{2} - 1)}, \frac{\partial}{\partial y} z (x, y) = - \frac{2 y}{2 z (x, y) - 1}]]$

•	Extract the solution for $u_{y}$ and replace $z (x, y)$ with $z$ .

$eval (op ([1, 1], q_{4}), [z (x, y) = z])$

$u_{y} = \frac{3 y (- 2 y z + 2 z^{2} + y)}{(2 z - 1) (3 u^{2} - 1)}$

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