Question

Find all first-order partial derivatives. (a) \(f(x, y, z)=\log (x+y+2 z)\) (b) \(f(x, y, z)=x^{2}+3 x y z+2 x y\) (c) \(f(x, y, z)=x e^{y z}\) (d) \(f(x, y, z)=z+\sin x^{2} y\)

Short answer

Question: Find the first-order partial derivatives of the following functions: (a) \(f(x, y, z)=\log (x+y+2z)\) (b) \(f(x, y, z)=x^{2}+3 x y z + 2xy\) (c) \(f(x, y, z)=x e^{yz}\) (d) \(f(x, y, z)=z+\sin x^{2}y\) Answer: (a) \(\frac{\partial f}{\partial x} = \frac{1}{x+y+2z}\), \(\frac{\partial f}{\partial y} = \frac{1}{x+y+2z}\), \(\frac{\partial f}{\partial z} = \frac{2}{x+y+2z}\) (b) \(\frac{\partial f}{\partial x} = 2x + 3yz + 2y\), \(\frac{\partial f}{\partial y} = 3xz + 2x\), \(\frac{\partial f}{\partial z} = 3xy\) (c) \(\frac{\partial f}{\partial x} = e^{yz}\), \(\frac{\partial f}{\partial y} = xze^{yz}\), \(\frac{\partial f}{\partial z} = xye^{yz}\) (d) \(\frac{\partial f}{\partial x} = 2xy\cos(x^2y)\), \(\frac{\partial f}{\partial y} = x^2\cos(x^2y)\), \(\frac{\partial f}{\partial z} = 1\)

Step-by-step solution

(a) Find the partial derivatives of \(f(x, y, z)=\log (x+y+2z)\)

Step 1: Differentiate with respect to x: \(\frac{\partial f}{\partial x} = \frac{1}{x+y+2 z} \cdot \frac{\partial}{\partial x}(x+y+2 z) = \frac{1}{x+y+2z}\) Step 2: Differentiate with respect to y: \(\frac{\partial f}{\partial y} = \frac{1}{x+y+2z} \cdot \frac{\partial}{\partial y}(x+y+2 z) = \frac{1}{x+y+2z}\) Step 3: Differentiate with respect to z: \(\frac{\partial f}{\partial z} = \frac{1}{x+y+2z} \cdot \frac{\partial}{\partial z}(x+y+2 z) = \frac{2}{x+y+2z}\)

(b) Find the partial derivatives of \(f(x, y, z)=x^{2}+3 x y z + 2xy\)

Step 1: Differentiate with respect to x: \(\frac{\partial f}{\partial x} = 2x + 3yz + 2y\) Step 2: Differentiate with respect to y: \(\frac{\partial f}{\partial y} = 3xz + 2x\) Step 3: Differentiate with respect to z: \(\frac{\partial f}{\partial z} = 3xy\)

(c) Find the partial derivatives of \(f(x, y, z)=x e^{yz}\)

Step 1: Differentiate with respect to x: \(\frac{\partial f}{\partial x} = e^{yz}\) Step 2: Differentiate with respect to y: \(\frac{\partial f}{\partial y} = x e^{yz} \cdot \frac{\partial}{\partial y}(yz) = xze^{yz}\) Step 3: Differentiate with respect to z: \(\frac{\partial f}{\partial z} = x e^{yz} \cdot \frac{\partial}{\partial z}(yz) = xye^{yz}\)

(d) Find the partial derivatives of \(f(x, y, z)=z+\sin x^{2}y\)

Step 1: Differentiate with respect to x: \(\frac{\partial f}{\partial x} = 0 + \cos(x^2 y) \cdot \frac{\partial}{\partial x}(x^2 y) = 2xy\cos(x^2y)\) Step 2: Differentiate with respect to y: \(\frac{\partial f}{\partial y} = 0 + \cos(x^2 y) \cdot \frac{\partial}{\partial y}(x^2 y) = x^2\cos(x^2y)\) Step 3: Differentiate with respect to z: \(\frac{\partial f}{\partial z} = 1 + 0 = 1\)

Key concepts

First-order derivatives

First-order derivatives are the building blocks of calculus. They show how a function changes as its input values change, specifically for a small increase or decrease. For a function of one variable, the first derivative with respect to that variable gives us the slope of the tangent line at any point. This gives us crucial information about the behavior of the function, such as whether it's increasing or decreasing.

In the context of partial derivatives, we extend these ideas to functions with multiple variables. Here, the first order derivative (called a partial derivative) with respect to one of the variables indicates how the function changes when only that variable is altered, keeping all others constant.

- **Example:** Consider the function (a) given, Continue with the differentiation process to find: 1. With respect to \(x\): \[ \frac{\partial f}{\partial x} = \frac{1}{x+y+2z} \] 2. With respect to \(y\): \[ \frac{\partial f}{\partial y} = \frac{1}{x+y+2z} \] 3. With respect to \(z\): \[ \frac{\partial f}{\partial z} = \frac{2}{x+y+2z} \]
These partial derivatives tell us how the function reacts to small changes in each of the variables \(x\), \(y\), and \(z\). They are fundamental to understanding the behavior of multivariable functions.

Multivariable calculus

Multivariable calculus extends the principles of calculus to functions of several variables. This is crucial in fields where phenomena are influenced by more than one factor, such as in physics, engineering, and economics. Unlike single-variable calculus, multivariable calculus works in higher dimensions and finds its applications in real-world problems.

When dealing with functions of three variables, such as \(f(x, y, z)\), we often compute partial derivatives to understand how the function behaves as each input changes independently. These derivatives are components of the gradient vector, which provides the direction of steepest ascent or descent of the function.

In exercise (b), dealing with the function \(f(x, y, z) = x^2 + 3xyz + 2xy\): - The partial derivative with respect to \(x\) is: \[ \frac{\partial f}{\partial x} = 2x + 3yz + 2y \] - With respect to \(y\): \[ \frac{\partial f}{\partial y} = 3xz + 2x \] - With respect to \(z\): \[ \frac{\partial f}{\partial z} = 3xy \]
This illustrates how changing each variable while holding the others steady affects the function's output. Such analyses are achieved through multivariable calculus concepts, enabling us to tackle more complex, real-world optimization problems.

Differentiation rules

Differentiation rules are fundamental to finding derivatives easily and accurately. In calculus, there are specific rules that help in determining derivatives of different types of functions, like products or compositions. For partial derivatives in multivariable calculus, these rules are still applicable but adjusted for several variables.

For example, some of the differentiation rules include:

• **Sum Rule**: The derivative of a sum of functions is the sum of their derivatives.
• **Product Rule**: Useful when differentiating products of functions. The derivative of \(u(x)\cdot v(x)\) is \(u'(x)\cdot v(x) + u(x)\cdot v'(x)\).
• **Chain Rule**: Allows for finding the derivative of composite functions.

Let's see these rules in action for the function \(f(x, y, z) = x e^{yz}\) given in exercise (c): - **Partial derivative with respect to \(x\):** Here, only \(e^{yz}\) is involved, resulting in: \[ \frac{\partial f}{\partial x} = e^{yz} \] - **Partial derivative with respect to \(y\):** Applying the product and chain rules involves differentiating \(e^{yz}\) with respect to \(y\), yielding: \[ \frac{\partial f}{\partial y} = xze^{yz} \] - **Partial derivative with respect to \(z\):** Similarly, differentiating \(e^{yz}\) with respect to \(z\) leads to: \[ \frac{\partial f}{\partial z} = xye^{yz} \]
These rules greatly simplify the process of differentiation, especially in problems involving multiple variables and complex expressions. Understanding and applying them accurately is key to mastering calculus.