Question

Find the first partial derivatives with respect to \(x, y,\) and \(z\). $$ H(x, y, z)=\sin (x+2 y+3 z) $$

Short answer

The first partial derivatives of the function \(H(x, y, z)\) with respect to \(x\), \(y\), and \(z\) are \(\frac{\partial H}{\partial x} = \cos(x + 2y + 3z)\), \(\frac{\partial H}{\partial y} = 2\cos(x + 2y + 3z)\), and \(\frac{\partial H}{\partial z} = 3\cos(x + 2y + 3z)\) respectively.

Step-by-step solution

Calculate the Partial Derivative with respect to x

The first partial derivative of \(H\) with respect to \(x\) is: \[ \frac{\partial H}{\partial x} = \cos(x + 2y + 3z) \cdot \frac{\partial}{\partial x} (x + 2y + 3z) = \cos(x + 2y + 3z) \cdot 1. \] Thus, \( \frac{\partial H}{\partial x} = \cos(x + 2y + 3z). \)

Calculate the Partial Derivative with respect to y

The first partial derivative of \(H\) with respect to \(y\) is: \[ \frac{\partial H}{\partial y} = \cos(x + 2y + 3z) \cdot \frac{\partial}{\partial y} (x + 2y + 3z) = \cos(x + 2y + 3z) \cdot 2. \] Thus, \( \frac{\partial H}{\partial y} = 2\cos(x + 2y + 3z). \)

Calculate the Partial Derivative with respect to z

The first partial derivative of \(H\) with respect to \(z\) is: \[ \frac{\partial H}{\partial z} = \cos(x + 2y + 3z) \cdot \frac{\partial}{\partial z} (x + 2y + 3z) = \cos(x + 2y + 3z) \cdot 3. \] Thus, \( \frac{\partial H}{\partial z} = 3\cos(x + 2y + 3z). \)

Key concepts

First Partial Derivative

Understanding the first partial derivative is essential for students delving into the realm of multivariable calculus. Imagine a surface that represents a function with two or more variables. In such a setting, when you want to find the rate of change of the function as you move along one of these variables, keeping the others constant, you calculate the first partial derivative with respect to that variable.

For instance, if we consider the function given in the exercise, \( H(x, y, z)=\sin (x+2y+3z) \), finding the first partial derivative with respect to \( x \) involves treating \( y \) and \( z \) as constants. This yields a new function that tells us how \( H \) changes as \( x \) varies, which is expressed as \( \frac{\partial H}{\partial x} = \cos(x + 2y + 3z) \). It's pivotal to grasp this concept to successfully analyze and interpret functions that depend on multiple variables.

Multivariable Calculus

Multivariable calculus is the branch of calculus that deals with functions of more than one variable. As opposed to single-variable calculus, where we work with functions like \( f(x) \) that depend on one variable, multivariable functions, such as \( H(x, y, z) \) from our exercise, depend on two or more variables. These functions can represent complex phenomena such as temperature changes over an area or the profitability of a product depending on different factors.

One of the pivotal tools in multivariable calculus is the concept of partial derivatives. When studying how a multivariable function changes in response to changes in one of its variables, partial derivatives come into play. Mastery of first partial derivatives leads to a deeper understanding of gradient vectors, optimization, and even the integral calculus of multiple variables.

Chain Rule in Calculus

The chain rule is a powerful tool in calculus, used to differentiate composite functions. When dealing with multivariable calculus, the chain rule extends to partial derivatives. It allows us to compute the derivative of a function that is composed of other functions, which are dependent on multiple variables.

The beauty of the chain rule lies in its ability to break down complex derivative problems into simpler parts. In the context of multivariable functions, when a variable within the function is itself a function of other variables, the chain rule helps in taking derivatives with respect to these interdependent variables. In simpler terms, it's like untangling a knot by addressing one loop at a time. For example, if we had a function \( G \) dependent on \( x \) and \( y \), where \( y \) itself is a function of \( x \), the chain rule would provide a systematic way to find \( \frac{dG}{dx} \), considering the indirect relationship through \( y \).