Problem 60 - Precalculus

Question

Refer to the following: In calculus, we find the derivative, \(f^{\prime}(x),\) of a function \(f(x)\) by allowing \(h\) to approach 0 in the difference quotient \(\frac{f(x+h)-f(x)}{h}\) of functions involving exponential functions. Find the difference quotient of \(f(x)=\cosh x\) and use it to prove that \((\cosh x)^{\prime}=\sinh x\).

Step-by-Step Solution

Answer

The derivative of \(\cosh x\) is \(\sinh x\).

Step 1: Difference Quotient Setup

The difference quotient of a function \(f(x)\) is given by \(\frac{f(x+h) - f(x)}{h}\). For \(f(x) = \cosh x\), this becomes \(\frac{\cosh(x+h) - \cosh x}{h}\).

Step 2: Expand \(\cosh(x+h)\)

Using the hyperbolic identity \(\cosh(x+h) = \cosh x \cdot \cosh h + \sinh x \cdot \sinh h\). Substitute this into the difference quotient: \[\frac{(\cosh x \cdot \cosh h + \sinh x \cdot \sinh h) - \cosh x}{h}\]

Step 3: Simplify the Expression

Now, simplify the expression by factoring out \(\cosh x\) and rearranging: \[\frac{\cosh x (\cosh h - 1) + \sinh x \cdot \sinh h}{h}\]

Step 4: Evaluate Limits as \(h\to 0\)

Take the limit as \(h\to 0\). Recall:- \(\lim_{h\to 0} \frac{\cosh h - 1}{h} = 0\)- \(\lim_{h\to 0} \frac{\sinh h}{h} = 1\)Applying these limits results in:\[\lim_{h\to 0}\frac{\cosh x (\cosh h - 1) + \sinh x \cdot \sinh h}{h} = \sinh x\]

Step 5: Conclusion

Thus, the derivative of \(\cosh x\) is \(\sinh x\). Therefore, we have shown that \((\cosh x)^{\prime} = \sinh x\).

Key Concepts

Difference Quotient

The difference quotient is a fundamental concept in calculus, often used to find the derivative of a function. It essentially measures how a function changes as its input changes. If you have a function, say \( f(x) \), you compute its difference quotient as \( \frac{f(x+h) - f(x)}{h} \). The idea is to look at the average rate of change of the function over a tiny (and shrinking) interval \( h \).

For example, if we take the hyperbolic cosine function, \( f(x) = \cosh x \), its difference quotient becomes \( \frac{\cosh(x+h) - \cosh x}{h} \). As \( h \) approaches zero, this difference quotient turns into the derivative of the function.

So, the difference quotient doesn't just assess change but does so in a way that helps us find instantaneous rates of change or slopes of tangent lines - a crucial part of understanding the behavior of functions!

Hyperbolic Functions

Hyperbolic functions, similar in spirit to trigonometric functions, have unique applications in mathematics, physics, and engineering. They are defined using exponential functions and are essential for describing certain types of growth and decay processes.

Some basic hyperbolic functions include:

• \( \sinh x \): the hyperbolic sine function, defined as \( \frac{e^x - e^{-x}}{2} \)
• \( \cosh x \): the hyperbolic cosine function, defined as \( \frac{e^x + e^{-x}}{2} \)

These functions exhibit properties similar to sine and cosine but differ by using exponential expressions. Hyperbolic functions often appear in the study of hyperbolas, hence their name. The identity \( \cosh^2 x - \sinh^2 x = 1 \), closely resembling the Pythagorean identity, is a standard part of their characteristic.

Understanding these functions can give insight into a wide array of mathematical phenomena, from the shape of a hanging cable (catenary) to complex numbers' geometry.

Derivative of Hyperbolic Functions

The process of finding the derivative of hyperbolic functions follows the principles used for other types of functions. A derivative gives insight into how a function behaves by describing its instantaneous rate of change. For the hyperbolic functions, these derivatives are straightforward but carry interesting properties.

For instance, the derivative of \( \cosh x \) is \( \sinh x \). This result comes from calculating the limit of its difference quotient, \( \frac{\cosh(x+h) - \cosh x}{h} \), as \( h \to 0 \).

Key rules and limits employed in reaching this result include:

• \( \lim_{h\to 0} \frac{\cosh h - 1}{h} = 0 \)
• \( \lim_{h\to 0} \frac{\sinh h}{h} = 1 \)

These limits help simplify the expression and reveal the derivative, showing the beauty and connections hyperbolic functions have with exponential functions. Mastering these derivatives can lead to solving more complex calculus problems with greater ease.