Problem 28 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \underline{\phantom{xxx}} , \text { find the indicated derivative. } $$ $$ D_{x} e^{2 x^{2}-x} $$

Step-by-Step Solution

Answer

The derivative is \((4x - 1) e^{2x^2 - x}\).

Step 1: Identify the function and the context

We are given the function \( f(x) = e^{2x^2 - x} \) and we need to find its derivative with respect to \( x \). The exponential function \( e^{u} \), where \( u = 2x^2 - x \), requires the use of the chain rule for differentiation.

Step 2: Differentiate using the Chain Rule

The chain rule states that if you have a composite function \( f(g(x)) \), the derivative is \( f'(g(x)) \cdot g'(x) \). Here, \( f(u) = e^u \) and \( u(x) = 2x^2 - x \). The derivative of \( e^u \) with respect to \( u \) is \( e^u \). So we apply the chain rule: \[ D_x[e^{2x^2 - x}] = e^{2x^2 - x} \cdot \frac{d}{dx}(2x^2 - x) \]

Step 3: Differentiate the inner function

Find the derivative of the inner function \( u(x) = 2x^2 - x \). Differentiate \( 2x^2 \) to get \( 4x \), and differentiate \( -x \) to get \( -1 \). Therefore, the derivative \( u'(x) = 4x - 1 \).

Step 4: Apply the Chain Rule to find the Derivative

Using the results from Steps 2 and 3, apply the chain rule: \[ D_x[e^{2x^2 - x}] = e^{2x^2 - x} \cdot (4x - 1) \] Thus, \[ \text{Derivative } D_x e^{2x^2-x} = (4x - 1) e^{2x^2 - x} \].

Key Concepts

Derivatives

Derivatives are a fundamental concept in calculus, representing the rate of change of a function. They're essential for understanding how functions behave and change over time. When we calculate derivatives, we're essentially finding how sensitive a function is to changes in its inputs. This is crucial for many applications in physics, engineering, economics, and more. For a simple function like \( f(x) = 2x^2 - x \), finding the derivative involves applying basic rules like the power rule. The power rule states that \( \frac{d}{dx}[x^n] = nx^{n-1} \). So, for \( 2x^2 \), the derivative becomes \( 4x \), and for \( -x \), it becomes \( -1 \). By combining these, we get \( 4x - 1 \) as the derivative of \( u(x) = 2x^2 - x \). Understanding derivatives is about grasping these rules and knowing how to apply them to different kinds of functions. It helps us predict and understand the behavior of real-world systems.

Exponential Function

The exponential function, denoted by \( e^x \), is one of the most significant functions in mathematics due to its unique properties. It’s the function where the rate of growth is proportional to its current value, making it a model for exponential growth or decay seen in nature and finance. The exponential function is special because its derivative is the same as the function itself: \( \frac{d}{dx}[e^x] = e^x \). This means that it grows at a rate proportional to its current value, which is why you see it used in contexts like population growth, radioactive decay, and interest compounding. In the context of our exercise, we have a function like \( e^{2x^2 - x} \), where \( e \) is raised to a more complex function, not just \( x \). That's where the chain rule comes in.

Composite Function

Composite functions involve applying one function to the result of another function. They’re essentially functions made up of other functions. In our example, \( f(x) = e^{2x^2-x} \), we have a composite function because we are applying the exponential function to \( 2x^2-x \). Understanding composite functions is crucial because they allow us to handle and model complex relationships in a structured way. When dealing with derivatives of composite functions, the chain rule is your best tool. The chain rule provides a way to differentiate these functions by unraveling the composition step by step. You follow this process: differentiate the outer function while keeping the inner one unchanged, then multiply by the derivative of the inner function. This approach simplifies the process of differentiation for complex, layered functions, making it manageable to find derivatives of functions like \( e^{2x^2-x} \) efficiently.