Problem 40 - Calculus Early Transcendentals: Pearson New International Edition

Question

If \(f(x, y, z)=\left(x^{3}+y^{2}+z\right)^{4}\), find each of the following: (a) \(f_{x}(x, y, z)\) (b) \(f_{y}(0,1,1)\) (c) \(f_{z z}(x, y, z)\)

Step-by-Step Solution

Answer

(a) \(f_x(x, y, z) = 12x^2(x^3 + y^2 + z)^3\); (b) \(f_y(0,1,1) = 64\); (c) \(f_{zz}(x, y, z) = 12(x^3 + y^2 + z)^2\).

Step 1: Understand the function

The function given is \(f(x, y, z) = (x^3 + y^2 + z)^4\). This is a composition of functions, where an outer function \(g(t) = t^4\) is applied to an inner function \(h(x, y, z) = x^3 + y^2 + z\). We need to find partial derivatives of \(f\) with respect to \(x\), \(y\), and second derivative with respect to \(z\).

Step 2: Differentiate with respect to x

To find \(f_{x}(x, y, z)\), use the chain rule. Let \(t = x^3 + y^2 + z\), then \(f = t^4\). First, find \(\frac{df}{dt} = 4t^3\). The partial derivative of \(t\) with respect to \(x\) is \(\frac{\partial t}{\partial x} = 3x^2\). Thus, \(f_x(x, y, z) = 4(x^3 + y^2 + z)^3 \cdot 3x^2 = 12x^2 (x^3 + y^2 + z)^3\).

Step 3: Evaluate the partial derivative with respect to y at a specific point

To find \(f_{y}(0, 1, 1)\), first find the partial derivative of \(f\) with respect to \(y\) using the chain rule: \(f_y(x, y, z) = 4 (x^3 + y^2 + z)^3 \cdot 2y = 8y(x^3 + y^2 + z)^3\). Substitute \(x = 0\), \(y = 1\), and \(z = 1\) into this expression: \(f_y(0, 1, 1) = 8 \cdot 1 (0^3 + 1^2 + 1)^3 = 8 \cdot 2^3 = 64\).

Step 4: Differentiate with respect to z twice

To find the second partial derivative with respect to \(z\), first find \(f_{z}(x, y, z)\). Apply the chain rule: \(f_z(x, y, z) = 4 (x^3 + y^2 + z)^3\). Now apply the derivative again with respect to \(z\): \(f_{zz}(x, y, z) = \frac{\partial}{\partial z} \left(4 (x^3 + y^2 + z)^3\right) = 12(x^3 + y^2 + z)^2\).

Key Concepts

Chain Rule.

The chain rule is a fundamental differentiation technique in calculus. It helps us find the derivative of composite functions.
When a function can be seen as a composition of two or more functions, as in our example where we have \(f(x, y, z) = (x^3 + y^2 + z)^4\), the chain rule allows us to differentiate by taking the derivative of the outer function and multiplying it by the derivative of the inner function.
Here, first set \(t = x^3 + y^2 + z\) as the inner function, and \(g(t) = t^4\) as the outer function. To find the partial derivative with respect to \(x\), the chain rule teaches us to first get \(\frac{dg}{dt} = 4t^3\), then find \(\frac{\partial t}{\partial x} = 3x^2\).
We multiply these results to get the desired derivative \(f_x(x, y, z) = 12x^2 (x^3 + y^2 + z)^3\).
Understanding this process deepens our comprehension of how changes in one direction affect the function's overall behavior.

Multivariable Calculus.

Multivariable calculus extends the principles of differentiation and integration to functions with more than one variable.
This is crucial because many real-world applications involve functions of several variables, such as temperature depending on time and location.
In our example, \(f(x, y, z)\), multivariable calculus allows us to explore how alterations in \(x\), \(y\), and \(z\) individually and collectively influence the entire function.
Partial derivatives are key tools in multivariable calculus used to analyze how a function changes with respect to one particular variable while keeping others constant.
By understanding partial derivatives, we gain insights into the behavior of complex functions in multiple dimensions, making it easier to model and analyze real-world phenomena.

Differentiation Techniques.

Differentiation techniques are methods used to find derivatives, the measures of how a function changes as its input changes, specially focusing on their application to multivariable functions.
We encountered such techniques when finding \(f_y(x, y, z) = 8y(x^3 + y^2 + z)^3\).
Here, applying the chain rule is central, but we can also utilize other techniques like the product rule or quotient rule when needed.
Differentiation techniques are sometimes combined to break down complex functions into simpler parts for easy calculations.
For instance, once we understand the technique's steps using simple functions, applying them to tougher, multivariable functions becomes straightforward.
Practicing differentiation techniques strengthens our mathematical skills, enabling efficient solutions for intricate scenarios.

Higher-Order Derivatives.

Higher-order derivatives provide insights into a function's deeper properties, like how its rate of change itself changes.
In our exercise, we explore this by calculating \(f_{zz}(x, y, z)\), or the second partial derivative with respect to \(z\).
The computation \(f_{zz}(x, y, z) = 12(x^3 + y^2 + z)^2\) involves differentiating twice, which helps understand concavity and rates of acceleration, instead of just direction and speed of change.
Higher-order derivatives are essential in fields like physics for analyzing motion, as they relate to understanding acceleration and other dynamic attributes.
Comprehending higher-order derivatives equips us to handle complex mathematical challenges, modeling advanced concepts and trends within diverse domains.