Problem 6 - Thomas Calculus in SI Units

Question

Rewrite the expressions in terms of exponentials and simplify the results as much as you can. $$\sinh (2 \ln x)$$

Step-by-Step Solution

Answer

\(\sinh(2 \ln x) = \frac{x^4 - 1}{2x^2}\)

Step 1: Recall Hyperbolic Sine Definition

The hyperbolic sine of a variable, \(x\), is defined as \(\sinh(x) = \frac{e^x - e^{-x}}{2}\). We will use this to rewrite \(\sinh(2 \ln x)\).

Step 2: Substitute and Expand

Substitute \(y = 2 \ln x\) in the hyperbolic sine definition: \(\sinh(2 \ln x) = \frac{e^{2 \ln x} - e^{-2 \ln x}}{2}\).

Step 3: Simplify Exponentials

Use the logarithmic identity \(e^{\ln a} = a\) to simplify the expression. For \(e^{2 \ln x}\), it simplifies to \((e^{\ln x})^2 = x^2\).For \(e^{-2 \ln x}\), it simplifies to \(1/(e^{\ln x})^2 = 1/x^2\).

Step 4: Substitute Simplified Terms

After simplification, substitute back into the equation: \(\sinh(2 \ln x) = \frac{x^2 - \frac{1}{x^2}}{2}\).

Step 5: Simplify the Fraction

Combine into a single fraction: \(\frac{x^2 - \frac{1}{x^2}}{2} = \frac{x^4 - 1}{2x^2}\).

Key Concepts

Exponential Functions

Exponential functions are a fundamental part of calculus and algebra. They are equations involving numbers raised to a power. Instead of using a base like 2 or 10, the base is often the mathematical constant \(e\) (approximately equal to 2.71828), known as Euler's number, a key figure in mathematics.

• Exponential growth appears in population growth, compound interest, and radioactive decay.
• Mathematically, an exponential function is defined as \(f(x) = a \cdot e^{kx}\), where the value \(a\) is the initial amount and \(k\) is a constant.
• Understanding exponential functions is crucial, as they're applicable in both math and real-world contexts.
• In hyperbolic functions like \(\sinh(x)\), the exponential function forms the core, with the formula \(\sinh(x) = \frac{e^x - e^{-x}}{2}\). This suggests the use of both positive and negative exponentials.

In our example, \(\sinh(2 \ln x)\) taps into exponential concepts by substituting logarithmic values into this fundamental equation. By simplifying \(e^{2 \ln x}\) to \(x^2\), we're leveraging these exponential foundations.

Logarithmic Identities

Logarithms are the inverses of exponentials and are just as important in simplifying expressions involving exponentials. The identity \(e^{\ln a} = a\) is a main tool in transforming and simplifying complex exponential expressions.

• Logarithmic identities enable us to switch between exponential and logarithmic forms, making calculations more manageable.
• With \(\ln(x)\), we focus on the natural logarithm (log to the base \(e\)), commonly used in calculus.
• Using identities such as \(\ln(a^b) = b\ln(a)\) helps in dealing with powers and roots of numbers.

Application in Our Problem

In the given problem, we used the logarithmic identity \(e^{2 \ln x} = (e^{\ln x})^2 = x^2\).This key transformation helps reduce the original complicated expressions involving exponentials to simpler algebraic terms, which are easier to handle and visualize.

Expression Simplification

Expression simplification is the process of altering an expression to make it easier to understand or solve. It's an essential skill in mathematics, creating the foundation for solving equations and identifying patterns.

• The main goal is to reduce expressions into their simplest form without changing their value.
• It includes combining like terms, reducing fractions, and using identities (like multiplying or dividing by common factors).

Steps Followed in Simplification

In our hyperbolic function problem, we took several essential steps:- After rewriting \(\sinh(2 \ln x)\) using exponential functions, we simplified \(e^{2 \ln x}\) to \(x^2\), and \(e^{-2 \ln x}\) to \(1/x^2\).- We then combined terms into a single fraction \(\frac{x^2 - \frac{1}{x^2}}{2}\).- Finally, we simplified further to obtain \(\frac{x^4 - 1}{2x^2}\), a neat expression that's easier to work with and understand. Simplification helps in revealing a structure that's more elegant and often more insightful than the original form.