91Ó°ÊÓ

The Product Rule is a crucial technique in differential calculus. It enables us to find the derivative of a product of two functions. For two functions, say \( u(x) \) and \( v(x) \), the rule is expressed as:

• \((u \, v)' = u' \, v + u \, v'\)

Here, the notation \((uv)'\) represents the derivative of the product of \( u \) and \( v \). In the given problem, our task is to differentiate the function \(y = e^{ax} \sin bx\).

Here’s how you apply the product rule:

• Let \( u = e^{ax} \) and \( v = \sin bx\).
• Differentiate \( u \) to get \( u' = a e^{ax} \).
• Differentiate \( v \) to get \( v' = b \cos bx \).

By substituting these into the product rule formula, we derive:

• \( y' = a e^{ax} \sin bx + e^{ax} b \cos bx \)

This understanding of the product rule helps in solving derivatives involving products of functions, especially when involving exponential and trigonometric functions.

Exponential functions are characterized by constants raised to the power of a variable. These functions have the form \( e^{ax} \), where \( e \) is the base and \( a \) is a constant.

Key properties of exponential functions include:

• Growth: Exponential functions grow rapidly as \( x \) increases.
• Derivation: The derivative of \( e^{ax} \) is \( a e^{ax} \), making it straightforward to handle in calculus.

In our exercise, \( e^{ax} \) serves as part of the product that needs differentiating.

When differentiating, understanding that the derivative remains proportional to the function itself (multiplied by the constant \( a \)) simplifies the process. This property makes exponential functions unique and integral to solving many calculus problems. Recognizing patterns within these functions allows for effective application of different calculus rules like the product rule.

Trigonometric functions, such as \( \sin \) and \( \cos \), relate the angles of a triangle to the lengths of its sides. In calculus, these functions offer periodic behavior, a vital aspect when differentiating.

Here are important aspects of trigonometric functions in differentiation:

• Derivative of \( \sin x \): \( (\sin x)' = \cos x \)
• Derivative of \( \cos x \): \( (\cos x)' = -\sin x \)

In the problem, \( \sin bx \) is differentiated as part of the product.

The derivative becomes \( b \cos bx \), thanks to the chain rule application, since \( b \) is constant and acts as a multiplier. Understanding these derivatives aids in predicting how changes in \( x \) affect the trigonometric function's output. With their regular oscillations, trigonometric functions, when combined with exponential functions, lead to interesting derivative results, as seen in the original exercise.