Problem 6 - A Primer of Real Analysis

Question

If \(f(x)=\sinh (x)\) and \(g(x)=\cosh (x),\) show that $$ f^{\prime}(x)=\cosh (x) $$ and $$ g^{\prime}(x)=\sinh (x) $$

Step-by-Step Solution

Answer

The derivative of \( f(x) = \sinh(x) \) is \( \cosh(x) \), and the derivative of \( g(x) = \cosh(x) \) is \( \sinh(x) \).

Step 1: Recall the Definitions

First, let's recall the definitions of hyperbolic sine and hyperbolic cosine. The hyperbolic sine function, \( \sinh(x) \), is defined as \( \frac{e^x - e^{-x}}{2} \). The hyperbolic cosine function, \( \cosh(x) \), is defined as \( \frac{e^x + e^{-x}}{2} \).

Step 2: Differentiate f(x)

Now, we need to differentiate the function \( f(x) = \sinh(x) = \frac{e^x - e^{-x}}{2} \). The derivative of \( e^x \) is \( e^x \) and the derivative of \( e^{-x} \) is \( -e^{-x} \). Thus, \( f'(x) = \frac{d}{dx} \left( \frac{e^x - e^{-x}}{2} \right) = \frac{1}{2} (e^x + e^{-x}) \). This simplifies to \( \cosh(x) \).

Step 3: Differentiate g(x)

Next, we differentiate the function \( g(x) = \cosh(x) = \frac{e^x + e^{-x}}{2} \). Using similar differentiation rules, \( g'(x) = \frac{d}{dx} \left( \frac{e^x + e^{-x}}{2} \right) = \frac{1}{2} (e^x - e^{-x}) \). This simplifies to \( \sinh(x) \).

Step 4: Conclusion

From these calculations, we conclude that the derivative of \( f(x) = \sinh(x) \) is \( f'(x) = \cosh(x) \) and the derivative of \( g(x) = \cosh(x) \) is \( g'(x) = \sinh(x) \). Thus, we have shown the required relationships for derivatives.

Key Concepts

Hyperbolic Sine Function

The hyperbolic sine function, denoted as \( \sinh(x) \), is a mathematical function that shares similar properties to the regular sine function from trigonometry, but is defined in the realm of hyperbolic geometry. Its formula is given by: \[ \sinh(x) = \frac{e^x - e^{-x}}{2} \]. This equation involves the exponential functions \(e^x\) and \(e^{-x}\), which are the basis for many calculations in calculus.

• \( e^x \) grows exponentially with positive \(x\), while \( e^{-x} \) decreases rapidly as \(x\) increases.
• The combination of \(e^x\) and \(e^{-x}\) in \(\sinh(x)\) results in a smooth curve that continues indefinitely in both directions of the x-axis.

To find the derivative of \(\sinh(x)\), you analyze the effect on the growth rates of these exponential components. As detailed in the original solution, differentiating \(\sinh(x)\) involves applying basic differentiation rules to each component:\[ f'(x) = \frac{d}{dx} \left( \frac{e^x - e^{-x}}{2} \right) = \frac{1}{2} \left( e^x + e^{-x} \right) \].This derivative surprisingly reveals that \(f'(x) = \cosh(x)\), linking these two hyperbolic functions in a complementary relationship.

Hyperbolic Cosine Function

Just like its sine counterpart, the hyperbolic cosine function, represented as \( \cosh(x) \), is an essential function in hyperbolic geometry. It is defined by the equation: \[ \cosh(x) = \frac{e^x + e^{-x}}{2} \].This function, too, employs the exponential functions \(e^x\) and \(e^{-x}\) but combines them differently, adding rather than subtracting.

• This equation features a positive sum, reflecting a characteristic hyperbolic curve that remains above the x-axis and never touches it.
• \( \cosh(x) \) is always greater than or equal to 1 for any real number \(x\).

Differentiation of \(\cosh(x)\) involves similar steps in applying differentiation rules to the exponential components. By calculating the derivative: \[ g'(x) = \frac{d}{dx} \left( \frac{e^x + e^{-x}}{2} \right) = \frac{1}{2} (e^x - e^{-x}) \],we find that \(g'(x) = \sinh(x)\). This is a fascinating result, as it shows the intertwined nature of hyperbolic sine and cosine functions in calculus.

Calculus Step-by-Step Solutions

Understanding calculus, especially differentiation, becomes more intuitive when approached with step-by-step solutions. In this specific exercise, the task was to differentiate two hyperbolic functions, \( \sinh(x) \) and \( \cosh(x) \). The recipe for successful differentiation of function involves these steps:

• First, recognize and write down the exponential definition of the hyperbolic functions. This step is crucial as it translates the problem into a more calculable form.
• Apply the derivative to each exponential term separately. For \(e^x\), the derivative is straightforward, \(e^x\). For \(e^{-x}\), remember the chain rule results in a negative derivative, \(-e^{-x}\).
• Combine the derivatives respecting their original positional addition or subtraction as defined by your hyperbolic function.

These solutions illustrate that taking derivatives yields a result that mirrors a predictable pattern:

• The derivative of \(\sinh(x)\) transforms into \(\cosh(x)\).
• The derivative of \(\cosh(x)\) becomes \(\sinh(x)\).

Encountering these results in a structured manner not only helps in solving such exercises but also deepens one’s understanding of the interconnected nature of hyperbolic functions in calculus. This step-by-step approach aids in demystifying the calculus processes and enhances learning efficiency.