Problem 14 - Thomas Calculus in SI Units

Question

Find the derivative of \(y\) with respect to the appropriate variable. $$y=\frac{1}{2} \sinh (2 x+1)$$

Step-by-Step Solution

Answer

The derivative is \( \cosh(2x + 1) \).

Step 1: Identify the Function

We are given the function \( y = \frac{1}{2} \sinh(2x + 1) \). Our task is to find its derivative with respect to \( x \).

Step 2: Differentiate with Respect to x

To differentiate \( y = \frac{1}{2} \sinh(2x + 1) \) with respect to \( x \), use the chain rule. Note that the derivative of \( \sinh u \) with respect to \( u \) is \( \cosh u \).

Step 3: Apply the Constant Multiple Rule

The constant multiple rule states that \( \frac{d}{dx}[c \, f(x)] = c \,\frac{d}{dx}[f(x)] \) where \( c \) is a constant. Thus, the derivative of \( \frac{1}{2} \sinh(2x + 1) \) is \( \frac{1}{2} \) times the derivative of \( \sinh(2x + 1) \).

Step 4: Use the Chain Rule

Apply the chain rule: \( \frac{d}{dx} [\sinh(2x + 1)] = \cosh(2x + 1) \cdot \frac{d}{dx}[2x + 1] \). The derivative of \( 2x + 1 \) with respect to \( x \) is 2.

Step 5: Simplify the Expression

Combining all steps, the derivative becomes: \( \frac{1}{2} \cdot \cosh(2x + 1) \cdot 2 = \cosh(2x + 1) \).

Step 6: Finalize the Derivative

The derivative of \( y = \frac{1}{2} \sinh(2x + 1) \) with respect to \( x \) is \( \cosh(2x + 1) \).

Key Concepts

Hyperbolic Functions

Hyperbolic functions are similar to trigonometric functions but are based on hyperbolas rather than circles. Just like sine and cosine form the basis of trigonometry, the hyperbolic sine (\( \sinh \)) and hyperbolic cosine (\( \cosh \)) form the basis of hyperbolic functions.

The hyperbolic sine function (\( \sinh \)) is defined as:

• \( \sinh(x) = \frac{e^x - e^{-x}}{2} \)

The derivative of the hyperbolic sine function is the hyperbolic cosine function:

• \( \frac{d}{dx}[\sinh(x)] = \cosh(x) \)

Similarly, the hyperbolic cosine function (\( \cosh \)) is defined as:

• \( \cosh(x) = \frac{e^x + e^{-x}}{2} \)

And its derivative is:

• \( \frac{d}{dx}[\cosh(x)] = \sinh(x) \)

Understanding these basic derivatives is crucial when working with more complex expressions, especially when combining hyperbolic functions with other operations like the chain rule.

Chain Rule

The chain rule is an essential tool in calculus for differentiating composite functions. It tells us how to differentiate a function that is made up of two or more nested functions.

In the expression \( y = \frac{1}{2} \sinh(2x + 1) \), we have an outer function and an inner function:

• Outer function: \( \sinh(u) \)
• Inner function: \( 2x + 1 \)

The chain rule states that:

• \( \frac{d}{dx}[h(g(x))] = h'(g(x)) \, g'(x) \)

Applied to our problem:

• The derivative of the outer function \( \sinh(u) \) with respect to \( u \) is \( \cosh(u) \).
• The derivative of the inner function \( 2x + 1 \) with respect to \( x \) is 2.

Thus, the chain rule gives us:

• \( \frac{d}{dx}[\sinh(2x+1)] = \cosh(2x+1) \cdot 2 \)

This systematic approach ensures accuracy when differentiating even more complex nested functions.

Differentiation Techniques

Differentiation is the process of finding the derivative of a function, which is essentially the rate at which the function's value changes. Several techniques can be applied depending on the form of the function you’re working with.

In this exercise, we've used a combination of the constant multiple rule and the chain rule. Let's briefly explore these techniques:

• **Constant Multiple Rule**: If you have a function multiplied by a constant, the constant remains unaffected. The derivative is simply the constant multiplied by the derivative of the function. Mathematically, this is expressed as:\( \frac{d}{dx}[c \cdot f(x)] = c \cdot \frac{d}{dx}[f(x)] \).


• **Chain Rule**: As discussed in the previous section, is used for differentiating composite functions.

These techniques streamlined the process of finding the derivative of \( y = \frac{1}{2} \sinh(2x + 1) \). Knowing which rule or combination of rules to apply is key when working through differentiation problems. Practice with these techniques ensures confidence and expertise in handling derivatives of various functions.