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 \(-2x^{-3} \sinh x + x^{-2} \cosh x\).

Step 1: Apply the Product Rule

To find the derivative of the function \(y = x^{-2} \sinh x\), we use the product rule for differentiation. The product rule states that for two functions \(u(x)\) and \(v(x)\), the derivative \(D_{x}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)\). Here, \(u(x) = x^{-2}\) and \(v(x) = \sinh x\).

Step 2: Differentiate the First Function

Find the derivative of \(u(x) = x^{-2}\). Using the power rule, \(u'(x) = -2x^{-3}\).

Step 3: Differentiate the Second Function

Now find the derivative of \(v(x) = \sinh x\). The derivative \(v'(x) = \cosh x\), since the derivative of \(\sinh x\) is \(\cosh x\).

Step 4: Apply the Derivatives to the Product Rule

Using the results from Steps 2 and 3, apply them in the product rule: \(D_{x}[x^{-2} \sinh x] = (-2x^{-3})(\sinh x) + (x^{-2})(\cosh x)\).

Step 5: Simplify the Expression

Combine the terms found in Step 4:\(-2x^{-3} \sinh x + x^{-2} \cosh x\).This is the derivative of the given function.

Key Concepts

Product Rule

The product rule is a fundamental tool in calculus used to differentiate products of two functions. It's particularly useful when you encounter complex expressions where simple rules don't suffice. In this exercise, we needed to differentiate the function \( y=x^{-2} \sinh x \). To do this efficiently, the product rule became necessary because our function is a product of two separate functions: \( u(x) = x^{-2} \) and \( v(x) = \sinh x \).

The rule states that if you have two differentiable functions \( u(x) \) and \( v(x) \), their derivative is given by:

• \( D_{x}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \)

This formula ensures that we accurately account for both functions changing simultaneously. In our exercise, each function was handled separately:

• First, we found \( u'(x) = -2x^{-3} \), the derivative of the first term.
• Second, we calculated \( v'(x) = \cosh x \), the derivative of the hyperbolic sine function.
• Lastly, we combined these into the product rule's format to find the final derivative.

Carrying out each step ensures accuracy and ease when managing more complicated differentiation exercises.

Differentiation

Differentiation is the process in calculus used to find the rate at which a function is changing at any given point. It mostly involves finding derivatives, which are essentially slopes of functions at specific points. This is a critical skill for understanding how behaviors of graphs change, for instance.

For any function, the derivative gives valuable insight into its growth and decay. Differentiation rules like the product rule, power rule, and chain rule, each serve to handle different mathematical scenarios.

• In this exercise, the power rule was utilized to differentiate \( u(x) = x^{-2} \).
• The power rule is simple: for \( x^n \), it yields \( nx^{n-1} \), and applied as \( u'(x) = -2x^{-3} \).
• In addition, we differentiated \( v(x) = \sinh x \) using the standard knowledge of hyperbolic derivatives, resulting in \( v'(x) = \cosh x \).

Differentiation, with its rules, forms the backbone of calculus, giving analysts and mathematicians an efficient toolkit for tackling complex functions.

Hyperbolic Functions

Hyperbolic functions, much like trigonometric ones, are special functions important in various mathematical and engineering contexts. They arise naturally in problems of hyperbolic geometry as well as many real-world problems involving calculations similar to trigonometric functions, but with different properties.

Two fundamental hyperbolic functions are \( \sinh x \) (hyperbolic sine) and \( \cosh x \) (hyperbolic cosine). These functions are defined as:

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

These resemble \( \sin x \) and \( \cos x \) from trigonometry but exhibit exponential characteristics.

In this problem, the differentiation of \( \sinh x \) yielded \( \cosh x \) due to their unique property where the derivative of \( \sinh x \) is \( \cosh x \). Understanding these concepts not only helps with this exercise but also broadens one's capability to handle complex mathematical equations within and beyond calculus.