Problem 17 - Thomas Calculus

Question

In Exercises \(17-36,\) find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$ y=e^{-5 x} $$

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dx} = -5e^{-5x} \).

Step 1: Recognize the Function Form

The given function is an exponential function: \( y = e^{-5x} \). We recognize this as \( y = e^{u} \) with \( u = -5x \).

Step 2: Determine the Derivative Formula

For exponential functions of the form \( y = e^{u} \), where \( u \) is a function of \( x \), the derivative with respect to \( x \) is given by \( \frac{dy}{dx} = e^{u} \cdot \frac{du}{dx} \).

Step 3: Compute \( \frac{du}{dx} \)

Since \( u = -5x \), we find its derivative with respect to \( x \): \( \frac{du}{dx} = -5 \).

Step 4: Apply the Derivative Formula

Substitute \( e^{u} = e^{-5x} \) and \( \frac{du}{dx} = -5 \) into the derivative formula: \( \frac{dy}{dx} = e^{-5x} \cdot (-5) \).

Step 5: Simplify the Expression

Simplify the expression to get the derivative: \( \frac{dy}{dx} = -5e^{-5x} \).

Key Concepts

Exponential Function

An exponential function is a mathematical expression where a constant base is raised to a variable power. In our exercise, we have the function \( y = e^{-5x} \), where \( e \) is the constant base. The number \( e \) is known as Euler's number, approximately equal to 2.71828. This number is unique because the function \( e^x \) has a distinctive property: its rate of change at any point is equal to its value at that point.

Exponential functions come in various forms, but a pure form always has the constant \( e \) raised to some expression involving a variable like \( x \).

• These functions model continuous growth or decay processes, such as population growth or radioactive decay, where the rate of change depends on the current amount.
• In our equation, \( y = e^{-5x} \), the exponent \(-5x\) implies a decay process because of the negative sign in front of the exponent.

Recognizing an exponential function is crucial for determining how to differentiate it, which is the next step in our exercise.

Derivative Formula

Differentiation is the process of finding the derivative of a function, which represents the function's rate of change. When dealing with exponential functions of the form \( y = e^{u} \) where \( u \) is a function of \( x \), we use a specific derivative formula:

• The formula is \( \frac{dy}{dx} = e^{u} \cdot \frac{du}{dx} \).
• Here, \( \frac{du}{dx} \) is the derivative of the inner function \( u \) with respect to \( x \).

For the function \( y = e^{-5x} \), we identify \( u = -5x \) and calculate \( \frac{du}{dx} = -5 \). This simplifies our differentiation process to finding \( \frac{dy}{dx} \) by multiplying the function itself by the derivative of its exponent:

• Hence, \( \frac{dy}{dx} = e^{-5x} \cdot (-5) \).
• This gives the final simplified derivative: \( \frac{dy}{dx} = -5e^{-5x} \).

Understanding this formula allows us to easily differentiate complex exponential functions by breaking them down into manageable parts.

Chain Rule

The chain rule is a fundamental principle in calculus used to differentiate composite functions. A composite function is a function within another function, denoted as \( y = f(g(x)) \). The derivative \( \frac{dy}{dx} \) can be found using the chain rule:

• It states that \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).
• This involves differentiating the outer function evaluated at the inner function and multiplying by the derivative of the inner function.

In the context of our problem, the chain rule applies to the differentiation of \( y = e^{-5x} \). Here, the outer function is \( e^{u} \), and the inner function is \( u = -5x \). When using the chain rule:

• We first differentiate the outer function, getting \( e^{u} \), which stays as \( e^{-5x} \) in our case.
• Then, we multiply by the derivative of the inner function, \( \frac{du}{dx} = -5 \).

Therefore, the chain rule leads us to the derivative \( \frac{dy}{dx} = -5e^{-5x} \). This process helps handle derivatives in a straightforward manner, especially when dealing with more complex nested functions.