Question

Let \(w=x^{2}+x y+z^{2}\) (a) If \(x^{3}+x=3 t, y^{4}+y=4 t, z^{5}+z=5 t,\) find \(d w / d t\) (b) If \(y^{3}+x y=1\) and \(z^{3}-x z=2,\) find \(d w / d x\) (c) If \(x^{3} z+z^{3} y+y^{3} x=0,\) find \((\partial w / \partial x)_{y}\).

Short answer

(a) \( \frac{dw}{dt} = (2x+y)\frac{3}{3x^2+1} + x\frac{4}{4y^3+1} + 2z\frac{5}{5z^4+1} \) (b) \( \frac{dw}{dx} = 2x + y + x\frac{-y}{3y^2 + x} + 2z\frac{z}{3z^2 - x} \) (c) \( \frac{\text{d} w}{\text{d} x}_{|y} = 2x + y + y^3 - 3x^2z \)

Step-by-step solution

Understanding the given function

The given function is \( w = x^2 + xy + z^2 \). We need to determine several derivatives based on additional given equations.

Calculating \( \frac{d w}{d t} \)

To find \( \frac{d w}{d t} \), apply the chain rule: \( \frac{d w}{d t} = \frac{\frac{\text{d}}{\text{d} x}(w)}{\frac{\text{d} x}{\text{d} t}} + \frac{\frac{\text{d}}{\text{d} y}(w)}{\frac{\text{d} y}{\text{d} t}} + \frac{\frac{\text{d}}{\text{d} z}(w)}{\frac{\text{d} z}{\text{d} t}} \).

Computing partial derivatives for \( \frac{d w}{d t} \)

Next compute \( \frac{\text{d}w}{\text{d}x} = 2x + y \), \( \frac{\text{d}w}{\text{d}y} = x \), and \( \frac{\text{d}w}{\text{d}z} = 2z \).

Finding \( \frac{\text{d}x}{\text{d}t} \), \( \frac{\text{d}y}{\text{d}t} \), \( \frac{\text{d}z}{\text{d}t} \)

Given \( x^3 + x = 3t \), differentiate implicitly with respect to t to get \( (3x^2 + 1)\frac{dx}{dt} = 3 \). Thus, \( \frac{dx}{dt} = \frac{3}{3x^2 + 1} \). Do similarly for y and z: \( \frac{dy}{dt} = \frac{4}{4y^3 + 1} \) and \( \frac{dz}{dt} = \frac{5}{5z^4 + 1} \).

Combining Partial Derivatives

Use the chain rule to combine everything for \( \frac{dw}{dt} \): \( \frac{dw}{dt} = (2x+y)\frac{3}{3x^2+1} + x\frac{4}{4y^3+1} + 2z\frac{5}{5z^4+1} \).

Calculating \( \frac{d w}{d x} \)

Given \( y^3 + xy = 1 \) and \( z^3 - xz = 2 \), we'll implicitly differentiate both: \( 3y^2 \frac{dy}{dx} + y + x \frac{dy}{dx} = 0 \), so \( \frac{dy}{dx} = \frac{-y}{3y^2 + x} \). For z: \( 3z^2 \frac{dz}{dx} - z - x\frac{dz}{dx} = 0 \), simplify to \( \frac{dz}{dx} = \frac{z}{3z^2 - x} \).

Applying the Chain Rule for \( \frac{\text{d} w}{\text{d} x} \)

Using the chain rule, \( \frac{d w}{d x} = \frac{\text{d} w}{\text{d} x} + \frac{\text{d} w}{\text{d} y}\frac{\text{d} y}{\text{d} x} + \frac{\text{d} w}{\text{d} z}\frac{\text{d} z}{\text{d} x} \), Plugging known values: \( \frac{d w}{d x} = 2x + y + x\frac{-y}{3y^2 + x} + 2z\frac{z}{3z^2 - x} \).

Calculating \( \frac{\text{d} w}{\text{d} x}_{|y} \)

For part (c), given \( x^3z + z^3y + y^3x = 0 \), differentiate with respect to x while keeping y constant. Thus, \( 3x^2z + z^3 \frac{\text{d} y}{\text{d} x} + y^3 = 0 \). So, we compute the partial derivative \( \frac{\text{d} w}{\text{d} x} = 2x + y + y^3 - 3x^2z \).

Key concepts

Chain Rule for Differentiation

The chain rule is a fundamental principle when dealing with derivatives in calculus, particularly when functions are composed or implicitly involve other variables. For a multivariable function, if one variable depends on another, the chain rule helps compute the derivative by considering all paths through which the primary variable affects the other.
For instance, if we need to find \(\frac{dw}{dt}\) for a function \(w(x, y, z)\) where each of \(x, y, z\) depends on \(t\), the chain rule states: \[ \frac{dw}{dt} = \frac{\frac{\text{d} w}{\text{d} x}}{\frac{\text{d} x}{\text{d} t}} + \frac{\frac{\text{d} w}{\text{d} y}}{\frac{\text{d} y}{\text{d} t}} + \frac{\frac{\text{d} w}{\text{d} z}}{\frac{\frac{\text{d} z}{\text{d} t}}} \]
This formula integrates the direct rate of change of \(w\) with respect to \(t\) through each dependent variable. We first compute the partial derivatives \(\frac{d w}{d x}, \frac{d w}{d y} \), and \(\frac{d w}{d z}\) and then multiply each by the corresponding derivative of \(x, y, z\) with respect to \(t\). Combining these components gives us \(\frac{d w}{d t}\).
Example: Given \( w = x^2 + xy + z^2 \), and the dependencies \( x^3 + x = 3t \), \( y^4 + y = 4t \), and \( z^5 + z = 5t \). We use implicit differentiation to find: \( \frac{dx}{dt} = \frac{3}{3x^2 + 1} \), \( \frac{dy}{dt} = \frac{4}{4y^3 + 1} \), and \( \frac{dz}{dt} = \frac{5}{5z^4 + 1} \). Combining these, the chain rule gives \( \frac{dw}{dt} = (2x + y)\frac{3}{3x^2 + 1} + x\frac{4}{4y^3 + 1} + 2z\frac{5}{5z^4 + 1} \).

Implicit Differentiation

Implicit differentiation is a technique used when a function is defined implicitly rather than explicitly. In contrast to explicit functions where y is given directly in terms of x (i.e., \(y = f(x)\)), implicit functions involve equations where y and x are intertwined (e.g., \(y^3 + xy = 1\)).
To differentiate implicitly:

• Treat all variables as functions of a single independent variable, often x or t.
• Differentiate both sides of the equation with respect to the chosen independent variable.
• Solve for the derivative of the dependent variable.

For instance, given \(y^3 + xy = 1\) and \(z^3 - xz = 2\), to find \(\frac{dw}{dx}\), we perform implicit differentiation: \[y^3 + xy = 1 \, \frac{d}{dx}(y^3 + xy) = \frac{d}{dx}(1) \], leading to: \[ 3y^2 \frac{dy}{dx} + y + x \frac{dy}{dx} = 0 \]. Solving for \(\frac{dy}{dx}\), we get \[ \frac{dy}{dx} = \frac{-y}{3y^2 + x} \]. Similarly, for \(z^3 - xz = 2\), differentials yield: \[ 3z^2 \frac{dz}{dx} - z - x \frac{dz}{dx} = 0 \, \frac{dz}{dx} = \frac{z}{3z^2 - x} \].
With these results, the chain rule provides a complete expression for \( \frac{d w}{d x} = 2x + y + x(\frac{-y}{3y^2 + x}) + 2z(\frac{z}{3z^2 - x}) \).

Multivariable Functions

Multivariable functions involve more than one variable and are a core concept in calculus, particularly when dealing with real-world phenomena. These functions can represent surfaces or three-dimensional shapes, where each input pair or triplet affects the output.
For example, the function \(w = x^2 + xy + z^2\) takes three variables (x, y, z) as inputs, and any change in these variables impacts the value of \(w\). When working with such functions, partial derivatives become essential tools.

• Partial Derivatives: They measure the rate of change of the function with respect to one variable while holding others constant. For \( w = x^2 + xy + z^2 \), the partial derivatives are: \(\frac{\partial w}{\partial x} = 2x + y\), \( \frac{\partial w}{\partial y} = x \), and \(\frac{\partial w}{\partial z} = 2z \).
• Directional Derivatives: Extend the concept of partial derivatives to find the rate of change in any direction.

The understanding of multivariable functions is critical in fields ranging from physics to economics, where models often depend on multiple variables.
To compute \(\frac{\partial w}{\partial x}\), given equations like \(x^3 z + z^3 y + y^3 x = 0\), partially differentiate concerning x while keeping other variables constant. Thus: \[ 3x^2 z + z^3 (\frac{dy}{dx}) + y^3 = 0 \, \frac{\partial w}{\partial x} = 2x + y + y^3 - 3x^2z \].