Problem 21 - Calculus Early Transcendentals

Question

Derive the following derivative formulas given that \(d / d x(\cosh x)=\sinh x\) and \(d / d x(\sinh x)=\cosh x\). \(d / d x(\operatorname{csch} x)=-\operatorname{csch} x \operatorname{coth} x\)

Step-by-Step Solution

Answer

Question: Find the derivative of \(y = \operatorname{csch} x\). Answer: The derivative of \(y = \operatorname{csch} x\) with respect to \(x\) is given by: \(\frac{d y}{d x} = -\operatorname{csch}^2 x \cdot \operatorname{coth} x\).

Step 1: Define the function

Given that \(\operatorname{csch} x = \frac{1}{\sinh x}\), we can express this as \(y=\operatorname{csch} x\) and rewrite it using the definition of \(\operatorname{csch} x\): $$ y = \frac{1}{\sinh x} $$

Step 2: Use the chain rule

To find the derivative of \(y\), we will use the chain rule, which states that: $$ \frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x} $$ Here, we'll let \(u = \sinh x\). Thus: $$ y = \frac{1}{u} $$ Now, we will find the derivatives of \(y\) with respect to \(u\) and \(u\) with respect to \(x\).

Step 3: Differentiate with respect to \(u\)

To find \(\frac{d y}{d u}\), differentiate \(y = \frac{1}{u}\) with respect to \(u\): $$ \frac{d y}{d u} = -\frac{1}{u^2} $$

Step 4: Find the derivative of \(u\) with respect to \(x\)

Recall that \(u = \sinh x\) and, given that \(\frac{d }{d x}(\sinh x) = \cosh x\), we have: $$ \frac{d u}{d x} = \cosh x $$

Step 5: Apply chain rule

Now, substituting the results from Steps 3 and 4 back into the chain rule formula, we get: $$ \frac{d y}{d x} = -\frac{1}{u^2} \cdot \cosh x $$

Step 6: Substitute \(u\) back and simplify

Now, replace \(u\) with its original expression \(\sinh x\): $$ \frac{d y}{d x} = -\frac{1}{(\sinh x)^2} \cdot \cosh x $$ Recall the definition of \(\operatorname{csch} x\) and $\operatorname{coth} x$ which are given by: $$ \operatorname{csch} x = \frac{1}{\sinh x} $$ and $$ \operatorname{coth} x = \frac{\cosh x}{\sinh x} $$ We can rewrite the derivative in terms of these functions: $$ \frac{d y}{d x} = -\operatorname{csch}^2 x \cdot \operatorname{coth} x $$ This expression represents the derivative of the function \(y = \operatorname{csch} x\).

Key Concepts

Derivative Formulas

Derivative formulas are fundamental tools in calculus, used to find the rate at which a function changes at any given point. Knowing how to derive formulas is crucial in understanding the behavior of various functions. For hyperbolic functions, the derivatives of basic functions like hyperbolic sine (\( \sinh x \)) and hyperbolic cosine (\( \cosh x \)) serve as building blocks for more complex derivatives.

In this problem, you're asked to find the derivative of the hyperbolic cosecant function, denoted as \( \operatorname{csch} x \). This is done through a series of steps that apply differentiation rules like the chain rule, combined with understanding hyperbolic functions.These formulas serve as tools to decompose problems into manageable parts and find derivatives efficiently.

• The derivative of \( \sinh x \) is \( \cosh x \), and vice versa.
• Through manipulation and substitution, derivatives of other composite hyperbolic functions are derived.

Applying these formulas systematically will enable you to tackle complex differentiation problems with confidence.

Chain Rule

The chain rule is a principle that helps in finding the derivative of composite functions. It states that if a function \( y \) depends on \( u \), and \( u \) depends on \( x \), the derivative of \( y \) with respect to \( x \) can be found by multiplying the derivative of \( y \) with respect to \( u \) by the derivative of \( u \) with respect to \( x \):\[\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}\]This rule is particularly useful when dealing with layers of functions, allowing you to break them down and differentiate each part step-by-step.

In the exercise, the chain rule was employed by setting \( u = \sinh x \), simplifying the expression \( y = \frac{1}{u} \) to tackle the problem methodically:

• First, find \( \frac{d y}{d u} \), the derivative of \( y \) with respect to \( u \).
• Then find \( \frac{d u}{d x} \), the derivative \( u \) with respect to \( x \).
• Combine these using the chain rule to find the desired derivative.

This systematic approach clarifies how the original function changes and helps avoid common pitfalls when dealing with nested expressions.

Cosh and Sinh Derivatives

The derivatives of hyperbolic cosine and sine, \( \cosh x \) and \( \sinh x \), are particularly straightforward: they directly relate to each other. These functions mirror, in many ways, the properties that sine and cosine have in trigonometry, but within the framework of hyperbolic geometry.

• The derivative of \( \cosh x \) is \( \sinh x \).
• Conversely, the derivative of \( \sinh x \) is \( \cosh x \).

Understanding these basic derivatives is essential, as they form the foundation for deriving more complex hyperbolic functions like \( \operatorname{csch} x \) and \( \operatorname{coth} x \).

In this problem, knowing these core derivatives allows you to express \( \operatorname{csch} x = \frac{1}{\sinh x} \). When differentiating \( \operatorname{csch} x \), it involves recognizing these relationships and applying them to find how the hyperbolic cosecant changes with respect to \( x \).This knowledge enables you to approach and solve function derivatives with confidence, ensuring a strong grasp on hyperbolic functions in calculus.