Problem 5 - Calculus for Biology and Medicine

Question

Differentiate the functions in Problems 1-52 with respect to the independent variable. $$ f(x)=e^{-2 x^{2}+3 x-1} $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = (-4x + 3) \cdot e^{-2x^2 + 3x - 1} \).

Step 1: Recall the Chain Rule

The Chain Rule in calculus is a formula to calculate the derivative of a composition of functions. If you have a function in the form of \( f(g(x)) \), then its derivative is calculated as \( f'(g(x)) \cdot g'(x) \).

Step 2: Identify the Inner Function

In the given function \( f(x) = e^{-2x^2 + 3x - 1} \), the inner function is the exponent part \( g(x) = -2x^2 + 3x - 1 \).

Step 3: Compute the Derivative of the Inner Function

Differentiate the inner function \( g(x) = -2x^2 + 3x - 1 \). This yields \( g'(x) = -4x + 3 \).

Step 4: Differentiate the Outer Function

The outer function is the exponential function \( f(g) = e^g \). The derivative of \( e^g \) with respect to \( g \) is \( e^g \).

Step 5: Apply the Chain Rule

Using the Chain Rule, the derivative of \( f(x) = e^{-2x^2 + 3x - 1} \) is \( f'(x) = e^{-2x^2 + 3x - 1} \cdot (-4x + 3) \).

Step 6: Simplify the Expression

Multiplying \( e^{-2x^2 + 3x - 1} \) by \(-4x + 3\) gives the final derivative: \( f'(x) = (-4x + 3) \cdot e^{-2x^2 + 3x - 1} \).

Key Concepts

Understanding the Chain Rule

The Chain Rule is an essential tool in calculus for differentiating compositions of functions. It allows us to find the derivative of a complex function by breaking it down into simpler parts. In simple terms, when we have a function that is made up of other functions, like \( f(g(x)) \), the Chain Rule tells us to:

• Differentiate the outer function while keeping the inner function unchanged.
• Multiply the result by the derivative of the inner function.

In our original exercise, the function is \( f(x) = e^{-2x^2 + 3x - 1} \), where the outer function is the exponential function, and the inner function is the exponent, \( g(x) = -2x^2 + 3x - 1 \). Applying the Chain Rule helps in evaluating the derivative by focusing on these individual components.
Using this technique ensures that you correctly differentiate the entire composition by respecting the role of each function involved. It is like peeling an onion: you work through the layers one at a time.

Derivative of Exponential Functions

Exponential functions, particularly those involving the natural exponential \( e \), have a unique property. The derivative of an exponential function \( e^g \) with respect to its variable \( g \) is simply \( e^g \) itself. This property makes calculations involving exponential functions both straightforward and quick.When differentiating the function \( f(x) = e^{-2x^2 + 3x - 1} \), identifying \( e^g \) as the natural exponential makes it easier to apply derivative rules. Notice how we keep the exponential part unchanged in the derivative process.
This is due to the special derivative property of exponential functions: the derivative of \( e^{\text{something}} \) is essentially \( e^{\text{same-something}} \) multiplied by the derivative of that "something."
In our example, the outer function's derivative is \( e^{-2x^2 + 3x - 1} \), which is multiplied by the derivative of the inner function as a result of applying the Chain Rule.
This continuous self-replicating nature of the exponential function’s derivative is an incredible feature that simplifies differentiation considerably.

Composition of Functions

The composition of functions involves creating a new function by applying one function to the results of another. For instance, if you have two functions, \( f(u) \) and \( g(x) \), their composition is expressed as \( f(g(x)) \). This composition forms the basis for applying the Chain Rule and is a common occurrence in calculus problems. For the function \( f(x) = e^{-2x^2 + 3x - 1} \), we have the function \( g(x) = -2x^2 + 3x - 1 \) nested inside the exponential function.
Understanding this layering allows us to logically and effectively use the Chain Rule by realizing that this multilayered function can be seen as a function nested within another.
By identifying a function as a composition, you significantly go forward in organizing your calculus strategy, handling each layer systematically.
This knowledge of dissecting a composition into its core components simplifies the differentiation process, making complex problems more manageable.