4.5 The chain rule (Page 4/9)

Calculus volume 3 Page 4 / 9

In the next example we calculate the derivative of a function of three independent variables in which each of the three variables is dependent on two other variables.

Using the generalized chain rule

Calculate $\partial w / \partial u$ and $\partial w / \partial v$ using the following functions:

\begin{matrix} w & = & f (x, y, z) = 3 x^{2} - 2 x y + 4 z^{2} \\ x & = & x (u, v) = e^{u} sin v \\ y & = & y (u, v) = e^{u} cos v \\ z & = & z (u, v) = e^{u} . \end{matrix}

The formulas for $\partial w / \partial u$ and $\partial w / \partial v$ are

\begin{matrix} \frac{\partial w}{\partial u} = \frac{\partial w}{\partial x} \cdot \frac{\partial x}{\partial u} + \frac{\partial w}{\partial y} \cdot \frac{\partial y}{\partial u} + \frac{\partial w}{\partial z} \cdot \frac{\partial z}{\partial u} \\ \frac{\partial w}{\partial v} = \frac{\partial w}{\partial x} \cdot \frac{\partial x}{\partial v} + \frac{\partial w}{\partial y} \cdot \frac{\partial y}{\partial v} + \frac{\partial w}{\partial z} \cdot \frac{\partial z}{\partial v} . \end{matrix}

Therefore, there are nine different partial derivatives that need to be calculated and substituted. We need to calculate each of them:

\begin{matrix} \frac{\partial w}{\partial x} & = & 6 x - 2 y & \frac{\partial w}{\partial y} & = & −2 x & \frac{\partial w}{\partial z} & = & 8 z \\ \frac{\partial x}{\partial u} & = & e^{u} sin v & \frac{\partial y}{\partial u} & = & e^{u} cos v & \frac{\partial z}{\partial u} & = & e^{u} \\ \frac{\partial x}{\partial v} & = & e^{u} cos v & \frac{\partial y}{\partial v} & = & − e^{u} sin v & \frac{\partial z}{\partial v} & = & 0 . \end{matrix}

Now, we substitute each of them into the first formula to calculate $\partial w / \partial u :$

\begin{matrix} \frac{\partial w}{\partial u} & = \frac{\partial w}{\partial x} \cdot \frac{\partial x}{\partial u} + \frac{\partial w}{\partial y} \cdot \frac{\partial y}{\partial u} + \frac{\partial w}{\partial z} \cdot \frac{\partial z}{\partial u} \\ = (6 x - 2 y) e^{u} sin v - 2 x e^{u} cos v + 8 z e^{u}, \end{matrix}

then substitute $x (u, v) = e^{u} sin v, y (u, v) = e^{u} cos v,$ and $z (u, v) = e^{u}$ into this equation:

\begin{matrix} \frac{\partial w}{\partial u} & = (6 x - 2 y) e^{u} sin v - 2 x e^{u} cos v + 8 z e^{u} \\ = (6 e^{u} sin v - 2 e^{u} cos v) e^{u} sin v - 2 (e^{u} sin v) e^{u} cos v + 8 e^{2 u} \\ = 6 e^{2 u} {sin}^{2} v - 4 e^{2 u} sin v cos v + 8 e^{2 u} \\ = 2 e^{2 u} (3 {sin}^{2} v - 2 sin v cos v + 4) . \end{matrix}

Next, we calculate $\partial w / \partial v :$

\begin{matrix} \frac{\partial w}{\partial v} & = \frac{\partial w}{\partial x} \cdot \frac{\partial x}{\partial v} + \frac{\partial w}{\partial y} \cdot \frac{\partial y}{\partial v} + \frac{\partial w}{\partial z} \cdot \frac{\partial z}{\partial v} \\ = (6 x - 2 y) e^{u} cos v - 2 x (− e^{u} sin v) + 8 z (0), \end{matrix}

then we substitute $x (u, v) = e^{u} sin v, y (u, v) = e^{u} cos v,$ and $z (u, v) = e^{u}$ into this equation:

\begin{matrix} \frac{\partial w}{\partial v} & = (6 x - 2 y) e^{u} cos v - 2 x (− e^{u} sin v) \\ = (6 e^{u} sin v - 2 e^{u} cos v) e^{u} cos v + 2 (e^{u} sin v) (e^{u} sin v) \\ = 2 e^{2 u} {sin}^{2} v + 6 e^{2 u} sin v cos v - 2 e^{2 u} {cos}^{2} v \\ = 2 e^{2 u} ({sin}^{2} v + sin v cos v - {cos}^{2} v) . \end{matrix}

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Calculate $\partial w / \partial u$ and $\partial w / \partial v$ given the following functions:

\begin{matrix} w & = & f (x, y, z) = \frac{x + 2 y - 4 z}{2 x - y + 3 z} \\ x & = & x (u, v) = e^{2 u} cos 3 v \\ y & = & y (u, v) = e^{2 u} sin 3 v \\ z & = & z (u, v) = e^{2 u} . \end{matrix}

$\begin{matrix} \frac{\partial w}{\partial u} = 0 \\ \frac{\partial w}{\partial v} = \frac{15 - 33 sin 3 v + 6 cos 3 v}{{(3 + 2 cos 3 v - sin 3 v)}^{2}} \end{matrix}$

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Drawing a tree diagram

Create a tree diagram for the case when

w = f (x, y, z), x = x (t, u, v), y = y (t, u, v), z = z (t, u, v)

and write out the formulas for the three partial derivatives of $w .$

Starting from the left, the function $f$ has three independent variables: $x, y, and z .$ Therefore, three branches must be emanating from the first node. Each of these three branches also has three branches, for each of the variables $t, u, and v .$

A diagram that starts with w = f(x, y, z). Along the first branch, it is written ∂w/∂x, then x = x(t, u, v), at which point it breaks into another three subbranches: the first subbranch says t and then ∂w/∂x ∂x/∂t; the second subbranch says u and then ∂w/∂x ∂x/∂u; and the third subbranch says v and then ∂w/∂x ∂x/∂v. Along the second branch, it is written ∂w/∂y, then y = y(t, u, v), at which point it breaks into another three subbranches: the first subbranch says t and then ∂w/∂y ∂y/∂t; the second subbranch says u and then ∂w/∂y ∂y/∂u; and the third subbranch says v and then ∂w/∂y ∂y/∂v. Along the third branch, it is written ∂w/∂z, then z = z(t, u, v), at which point it breaks into another three subbranches: the first subbranch says t and then ∂w/∂z ∂z/∂t; the second subbranch says u and then ∂w/∂z ∂z/∂u; and the third subbranch says v and then ∂w/∂z ∂z/∂v. — Tree diagram for a function of three variables, each of which is a function of three independent variables.

The three formulas are

\begin{matrix} \frac{\partial w}{\partial t} = \frac{\partial w}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial w}{\partial y} \frac{\partial y}{\partial t} + \frac{\partial w}{\partial z} \frac{\partial z}{\partial t} \\ \frac{\partial w}{\partial u} = \frac{\partial w}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial w}{\partial y} \frac{\partial y}{\partial u} + \frac{\partial w}{\partial z} \frac{\partial z}{\partial u} \\ \frac{\partial w}{\partial v} = \frac{\partial w}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial w}{\partial y} \frac{\partial y}{\partial v} + \frac{\partial w}{\partial z} \frac{\partial z}{\partial v} . \end{matrix}

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Create a tree diagram for the case when

w = f (x, y), x = x (t, u, v), y = y (t, u, v)

and write out the formulas for the three partial derivatives of $w .$

$\begin{matrix} \frac{\partial w}{\partial t} = \frac{\partial w}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial w}{\partial y} \frac{\partial y}{\partial t} \\ \frac{\partial w}{\partial u} = \frac{\partial w}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial w}{\partial y} \frac{\partial y}{\partial u} \\ \frac{\partial w}{\partial v} = \frac{\partial w}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial w}{\partial y} \frac{\partial y}{\partial v} \end{matrix}$

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Implicit differentiation

Recall from Implicit Differentiation that implicit differentiation provides a method for finding $d y / d x$ when $y$ is defined implicitly as a function of $x .$ The method involves differentiating both sides of the equation defining the function with respect to $x,$ then solving for $d y / d x .$ Partial derivatives provide an alternative to this method.

Consider the ellipse defined by the equation $x^{2} + 3 y^{2} + 4 y - 4 = 0$ as follows.

An ellipse with center near (0, –0.7), major axis horizontal and of length roughly 4.5, and minor axis of length roughly 3. — Graph of the ellipse defined by $x^{2} + 3 y^{2} + 4 y - 4 = 0 .$

This equation implicitly defines $y$ as a function of $x .$ As such, we can find the derivative $d y / d x$ using the method of implicit differentiation:

\begin{matrix} \frac{d}{d x} (x^{2} + 3 y^{2} + 4 y - 4) & = & \frac{d}{d x} (0) \\ 2 x + 6 y \frac{d y}{d x} + 4 \frac{d y}{d x} & = & 0 \\ (6 y + 4) \frac{d y}{d x} & = & −2 x \\ \frac{d y}{d x} & = & - \frac{x}{3 y + 2} . \end{matrix}

We can also define a function $z = f (x, y)$ by using the left-hand side of the equation defining the ellipse. Then $f (x, y) = x^{2} + 3 y^{2} + 4 y - 4 .$ The ellipse $x^{2} + 3 y^{2} + 4 y - 4 = 0$ can then be described by the equation $f (x, y) = 0 .$ Using this function and the following theorem gives us an alternative approach to calculating $d y / d x .$

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Source: OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2

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