Problem 28 - Essential Calculus Early Transcendentals

Question

Differentiate the function. $$ y=\frac{e^{u}-e^{-u}}{e^{u}+e^{-u}} $$

Step-by-Step Solution

Answer

The derivative is \( y' = \frac{4}{(e^u + e^{-u})^2} \).

Step 1: Apply the Quotient Rule

The formula for the derivative of a quotient is: \[\frac{d}{du}\left(\frac{f(u)}{g(u)}\right) = \frac{f'(u)g(u) - f(u)g'(u)}{[g(u)]^2}\]In our function, \(f(u) = e^{u} - e^{-u}\) and \(g(u) = e^{u} + e^{-u}\).

Step 2: Differentiate the Numerator

Now differentiate the numerator, \(f(u) = e^u - e^{-u}\). The derivative is: \[f'(u) = e^u + e^{-u}\]

Step 3: Differentiate the Denominator

Now differentiate the denominator, \(g(u) = e^u + e^{-u}\). The derivative is: \[g'(u) = e^u - e^{-u}\]

Step 4: Substitute Derivatives into the Quotient Rule

Substitute the derivatives into the quotient rule formula: \[y' = \frac{(e^u + e^{-u})(e^u + e^{-u}) - (e^u - e^{-u})(e^u - e^{-u})}{(e^u + e^{-u})^2}\]This simplifies to \[y' = \frac{(e^{2u} + 2 + e^{-2u}) - (e^{2u} - 2 + e^{-2u})}{(e^u + e^{-u})^2}\]which reduces to \[y' = \frac{4}{(e^u + e^{-u})^2}\]

Step 5: Simplify

Simplify the expression:\[y' = \frac{4}{(e^u + e^{-u})^2}\]The derivative can't be simplified further than this.

Key Concepts

Quotient Rule

When you have a function that is the division of two separate functions, you use the Quotient Rule to find its derivative. This tool in calculus helps differentiate functions that are expressed as one function divided by another. If you have a function

• \( y = \frac{f(u)}{g(u)} \)

then the Quotient Rule formula is:

• \( \frac{d}{du}\left(\frac{f(u)}{g(u)}\right) = \frac{f'(u)g(u) - f(u)g'(u)}{[g(u)]^2} \)

This formula tells you how to take the derivatives of both the numerator \( f(u) \) and the denominator \( g(u) \), and fit them into the expression.
The important thing to remember is:

• The minus in the numerator of the quotient rule, which results from multiplying the derivative of the whole function.
• You only apply the product rule to \( f'(u)g(u) - f(u)g'(u) \) in the numerator.
• Always square the denominator \( [g(u)]^2 \).

Hyperbolic Functions

Hyperbolic functions are similar to the trigonometric functions, but instead involve the exponential function \( e^u \). These functions appear in many applications across physics and engineering.
In differential calculus, the most important hyperbolic functions are hyperbolic sine \( \sinh(u) \) and hyperbolic cosine \( \cosh(u) \):

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

These functions are essential because they differentiate and integrate similarly to sine and cosine.
Often in problems, you end up working with these combinations of exponential terms - just like in our exercise:

• the numerator and denominator are actually multiples of hyperbolic functions.

Derivative Calculation

Performing derivative calculation requires you to carefully follow rules of differentiation - whether you are using basic rules or more complex ones like the Quotient Rule.
To differentiate a function, follow these broad steps:

• First identify the type of function you are dealing with, like rational (division) or trigonometric.
• Use applicable differentiation rules like the Quotient Rule or Product Rule, wherever necessary.
• Break down each function part by part. Differentiate the simpler, inner functions first.
• Substitute back differentiated parts into the required formula.
• Simplify the expression as much as possible to get the cleanest result.

In the example problem, the Quotient Rule helps simplify an initially complex derivative task into manageable steps that allow for easy simplification.
This ensures you always arrive at the simplest form of a derivative, like in our solution, where the derivative expression simplifies nicely to

• \( y' = \frac{4}{(e^u + e^{-u})^2} \).