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Chapter 4: Partial Differentiation

Section 4.3: Chain Rule

Example 4.3.27

If the equation $u = x + y + z + e^{u}$ implicitly defines $u = u (x, y, z)$ and the equation $z = x + y + \cos (z)$ 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.

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

and

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

from which it follows that $u_{x} = \frac{1 + (\frac{1}{1 + \sin (z)})}{1 - e^{u}} = \frac{2 + \sin (z)}{(1 - {&ExponentialE;}^{u}) (1 + \sin (z))}$ .

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

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

and

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

from which it follows that $u_{y} = \frac{1 + (\frac{1}{1 + \sin (z)})}{1 - e^{u}} = \frac{2 + \sin (z)}{(1 - {&ExponentialE;}^{u}) (1 + \sin (z))}$ .

Maple Solution - Interactive

Obtain $u_{x}$ from first principles

•	Write the first equation with the appropriate dependencies made explicit. Be sure to use Maple's exponential "e".

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

$u (x, y, z) = x + y + z (x, y) + {&ExponentialE;}^{u (x, y, z)}$

$u = x + y + z (x, y) + {&ExponentialE;}^{u}$

(1)

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

$u_{x} = 1 + \frac{\partial}{\partial x} z (x, y) + (u_{x}) {&ExponentialE;}^{u}$

(2)

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

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

$z (x, y) = x + y + \cos (z (x, y))$

(3)

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

$\frac{\partial}{\partial x} z (x, y) = 1 - (\frac{\partial}{\partial x} z (x, y)) \sin (z (x, y))$

(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 4.3.27(a).

Figure 4.3.27(a) Variables dialog

$,$

$u_{x} = 1 + \frac{\partial}{\partial x} z (x, y) + (u_{x}) {&ExponentialE;}^{u}, \frac{\partial}{\partial x} z (x, y) = 1 - (\frac{\partial}{\partial x} z (x, y)) \sin (z (x, y))$

(5)

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

$\{u_{x} = - \frac{2 + \sin (z (x, y))}{({&ExponentialE;}^{u} - 1) (\sin (z (x, y)) + 1)}, \frac{\partial}{\partial x} z (x, y) = \frac{1}{\sin (z (x, y)) + 1}\}$

(6)

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

Obtain $u_{y}$ from first principles

•	Write the first equation with the appropriate dependencies made explicit. Be sure to use Maple's exponential "e".

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

$u (x, y, z) = x + y + z (x, y) + {&ExponentialE;}^{u (x, y, z)}$

$u = x + y + z (x, y) + {&ExponentialE;}^{u}$

(7)

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

$u_{y} = 1 + \frac{\partial}{\partial y} z (x, y) + (u_{y}) {&ExponentialE;}^{u}$

(8)

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

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

$z (x, y) = x + y + \cos (z (x, y))$

(9)

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

$\frac{\partial}{\partial y} z (x, y) = 1 - (\frac{\partial}{\partial y} z (x, y)) \sin (z (x, y))$

(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 4.3.27(b).

Figure 4.3.27(b) Variables dialog

$,$

$u_{y} = 1 + \frac{\partial}{\partial y} z (x, y) + (u_{y}) {&ExponentialE;}^{u}, \frac{\partial}{\partial y} z (x, y) = 1 - (\frac{\partial}{\partial y} z (x, y)) \sin (z (x, y))$

(11)

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

$\{u_{y} = - \frac{2 + \sin (z (x, y))}{({&ExponentialE;}^{u} - 1) (\sin (z (x, y)) + 1)}, \frac{\partial}{\partial y} z (x, y) = \frac{1}{\sin (z (x, y)) + 1}\}$

(12)

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

Maple Solution - Coded

When writing the first equation, be sure to use Maple's exponential "e".

Apply the implicitdiff command

$implicitdiff (\{u = x + y + z + {&ExponentialE;}^{u}, z = x + y + \cos (z)\}, \{u, z\}, \{u, z\}, x)$

$\{D_{1} (u) = - \frac{2 + \sin (z)}{({&ExponentialE;}^{u} - 1) (\sin (z) + 1)}, D_{1} (z) = \frac{1}{\sin (z) + 1}\}$

$implicitdiff (\{u = x + y + z + {&ExponentialE;}^{u}, z = x + y + \cos (z)\}, \{u, z\}, \{u, z\}, y)$

$\{D_{2} (u) = - \frac{2 + \sin (z)}{({&ExponentialE;}^{u} - 1) (\sin (z) + 1)}, D_{2} (z) = \frac{1}{\sin (z) + 1}\}$

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

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. Be sure to use Maple's exponential "e". Press the Enter key.

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

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

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

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

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

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

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

$u_{x} = - \frac{2 + \sin (z)}{({&ExponentialE;}^{u} - 1) (\sin (z) + 1)}$

Obtain $u_{y}$

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

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

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

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

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

$u_{y} = - \frac{2 + \sin (z)}{({&ExponentialE;}^{u} - 1) (\sin (z) + 1)}$

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