Problem 10 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(D_{x} y\). $$ y=x^{-2} \sinh x $$

Step-by-Step Solution

Answer

The derivative is \( D_x y = -2x^{-3} \sinh x + x^{-2} \cosh x \).

Step 1: Identify the Function Type

The given function \( y = x^{-2} \sinh x \) is a product of two functions, namely \( u(x) = x^{-2} \) and \( v(x) = \sinh x \). Therefore, we will use the product rule to differentiate this function.

Step 2: Apply the Product Rule

The product rule states that if \( y = u(x) v(x) \), then \( D_x y = u'(x) v(x) + u(x) v'(x) \). We need to find the derivatives \( u'(x) \) and \( v'(x) \) separately.

Step 3: Differentiate \( u(x) = x^{-2} \)

Differentiate \( u(x) = x^{-2} \) with respect to \( x \). Use the power rule: \( u'(x) = -2x^{-3} \).

Step 4: Differentiate \( v(x) = \sinh x \)

Differentiate \( v(x) = \sinh x \) with respect to \( x \). The derivative of the hyperbolic sine function is the hyperbolic cosine function: \( v'(x) = \cosh x \).

Step 5: Substitute into the Product Rule Formula

Substitute \( u(x) = x^{-2} \), \( u'(x) = -2x^{-3} \), \( v(x) = \sinh x \), and \( v'(x) = \cosh x \) into the product rule:\[ D_x y = (-2x^{-3}) \sinh x + x^{-2} \cosh x \].

Step 6: Simplify the Expression

Combine the terms from the derivative:\[ D_x y = -2x^{-3} \sinh x + x^{-2} \cosh x \]. No further simplification is needed, and this is the final derivative.

Key Concepts

Product Rule in Derivative Calculus

To understand how to differentiate a product of two functions, we need to learn about the product rule. This is especially useful when tackling functions like the one in our exercise: \(y = x^{-2} \sinh x\). For such expressions, we use:

• Product Rule Formula: If \(y = u(x) v(x)\), then differentiating yields \(D_x y = u'(x) v(x) + u(x) v'(x)\).
• Function Identification: We have \(u(x) = x^{-2}\) and \(v(x) = \sinh x\).

By the product rule, each function is differentiated separately. We then combine the results, multiplying the derivative of each function by the other function in its original form. This helps in accurately capturing the change of two interacting parts in a product of functions.
Applying this rule allows us to evaluate the derivative such that both components of the product are accounted for properly.

Power Rule for Derivatives

The power rule is a fundamental technique for finding derivatives of functions of the form \(x^n\). In our example, we need it to differentiate \(u(x) = x^{-2}\). Here’s how the power rule works:

• Power Rule Formula: For any function \(f(x) = x^n\), the derivative is \(f'(x) = nx^{n-1}\).
• Application: With \(f(x) = x^{-2}\), this means \(f'(x) = -2x^{-3}\).

The power rule simplifies the process of differentiation by providing a quick method to find how a polynomial function's rate of change behaves. In our case, it tells us instantly what the derivative of \(x^{-2}\) is without messy calculations.
Whenever you see powers of \(x\), think of the power rule—it is a reliable and straightforward tool for differentiation that will serve you well.

Hyperbolic Functions in Derivatives

Hyperbolic functions, such as \(\sinh x\), resemble trigonometric functions but are defined using exponential functions. They include \(\sinh x\), \(\cosh x\), and others, each having unique properties.

• Differentiating \(\sinh x\): The derivative is \(\cosh x\), which is part of the special property of hyperbolic functions.
• Why It Matters: Recognizing the derivatives of hyperbolic functions is essential currently as they frequently appear in calculus problems involving complex and imaginary numbers.

These functions have their presence in many physical applications, from hyperbolic structures to calculus itself. Knowing that \(D_x[\sinh x] = \cosh x\) is part of the essential calculus toolkit, enabling you to handle problems effectively.
Whether they arise in product rules or stand-alone expressions, being prepared with the derivatives of hyperbolic functions can tremendously ease your journey through calculus.