Problem 126 - Calculus 3

Question

For the following exercises, calculate the partial derivatives. Let \(z=\sinh (2 x+3 y) .\) Find \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\)

Step-by-Step Solution

Answer

\(\frac{\partial z}{\partial x} = 2 \cosh(2x + 3y)\), \(\frac{\partial z}{\partial y} = 3 \cosh(2x + 3y)\).

Step 1: Identify the Function

Given the function \(z = \sinh(2x + 3y)\). This is a hyperbolic sine function where the argument is \(2x + 3y\). Our task is to find the partial derivatives, \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\).

Step 2: Differentiate with Respect to x

To find \(\frac{\partial z}{\partial x}\), apply the chain rule. The derivative of \(\sinh(u)\) is \(\cosh(u)\) where \(u = 2x + 3y\). The partial derivative becomes:\[\frac{\partial z}{\partial x} = \frac{d}{d(2x + 3y)}[\sinh(2x + 3y)] \cdot \frac{\partial}{\partial x}(2x + 3y) \]\[= \cosh(2x + 3y) \cdot 2\]Hence, the partial derivative of \(z\) with respect to \(x\) is \(\frac{\partial z}{\partial x} = 2 \cosh(2x + 3y)\).

Step 3: Differentiate with Respect to y

Now find \(\frac{\partial z}{\partial y}\) using a similar approach by applying the chain rule. We differentiate \(\sinh(u)\) with respect to \(u\) first, then multiply by \(\frac{\partial}{\partial y}(2x + 3y)\):\[\frac{\partial z}{\partial y} = \frac{d}{d(2x + 3y)}[\sinh(2x + 3y)] \cdot \frac{\partial}{\partial y}(2x + 3y)\]\[= \cosh(2x + 3y) \cdot 3\]Thus, the partial derivative of \(z\) with respect to \(y\) is \(\frac{\partial z}{\partial y} = 3 \cosh(2x + 3y)\).

Key Concepts

Hyperbolic Functions

Hyperbolic functions are closely related to the trigonometric functions but instead of the circle, they relate to the hyperbola. They appear in various areas of mathematics, physics, and even engineering, especially in topics involving the exponential function. The basic hyperbolic functions include \(\sinh\), \(\cosh\), and \(\tanh\).

• \(\sinh(x)\), which stands for hyperbolic sine, is defined as \((e^x - e^{-x})/2\).
• \(\cosh(x)\), known as hyperbolic cosine, is similarly defined by \((e^x + e^{-x})/2\).
• \(\tanh(x)\) is the hyperbolic tangent, representing the ratio of \(\sinh(x)\) to \(\cosh(x)\).

The key property that makes these functions useful is their relationship to the exponential function, providing smooth curves that switch between growth and decay rates. In the exercise, you encountered \(\sinh\) and its derivative \(\cosh\). This transformation is critical for partial differentiation as it simplifies complex expressions by using known derivative rules of hyperbolic functions.

Chain Rule

The chain rule is a fundamental tool in calculus, often used for differentiating compositions of functions. When dealing with hyperbolic functions, the chain rule helps us find the derivative of a compound function like \(\sinh(2x + 3y)\).

• First, identify the inner function, which in this example is \(u = 2x + 3y\).
• The chain rule states that the derivative of \(\sinh(u)\) is \(\cosh(u)\), and we must then multiply by the derivative of \(u\) with respect to our variable of interest.

This method provides a systematic way to tackle differentiation problems in multivariable contexts. It ensures that we account for changes that occur indirectly through the composite function's dependent variables. Hence, the chain rule is indispensable for working efficiently with complex functions in calculus.

Multivariable Calculus

Multivariable calculus expands the concepts of single-variable calculus to functions of several variables. It covers essential calculus tools like partial derivatives, multiple integrals, and vector calculus.In multivariable calculus, functions can depend on two or more variables, such as \(z = \sinh(2x + 3y)\). These kinds of functions often require more sophisticated analysis techniques than single-variable functions.

• Concepts such as gradient, divergence, and curl are tools from vector calculus used for analyzing multivariable functions.
• Functions are often visualized as surfaces instead of curves.
• Understanding how a function changes in response to several variables at once is crucial for fields like physics, economics, and engineering.

The exercise illustrates the application of multivariable calculus concepts as it involves functions with more than one input variable, motivated by real-world situations requiring a deeper understanding of varying relationships across dimensions.

Partial Differentiation

Partial differentiation is the process of finding the derivative of a function with respect to one variable while keeping all other variables constant. This technique is fundamental in analyzing functions that depend on multiple variables.

• In the exercise, you calculate partial derivatives \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\).
• For \(\frac{\partial z}{\partial x}\), treat \(y\) as constant and differentiate regarding \(x\).
• For \(\frac{\partial z}{\partial y}\), treat \(x\) as constant instead.

Partial differentiation provides insight into how a change in one particular direction can affect the outcome of a function. In fields like economics, it helps find marginal values, and in physics, it can describe change in multidimensional space. By learning partial differentiation, one gains a robust tool to tackle and understand multivariable problems.