Problem 11 - Thomas Calculus in SI Units

Question

Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$y=\left(x^{2}-2 x+2\right) e^{x}$$

Step-by-Step Solution

Answer

The derivative is \(y' = x^2 e^x\).

Step 1: Identify the Function Components

We are given the function \(y = (x^2 - 2x + 2) e^x\). It is a product of two functions: \(u = x^2 - 2x + 2\) and \(v = e^x\). To differentiate this expression, we will apply the product rule for derivatives.

Step 2: Apply the Product Rule for Derivatives

The product rule states that if \(y = uv\), then \(\frac{dy}{dx} = u'v + uv'\), where \(u'\) and \(v'\) are the derivatives of \(u\) and \(v\), respectively. Let's compute these derivatives next.

Step 3: Differentiate the First Function Component

For \(u = x^2 - 2x + 2\), differentiate with respect to \(x\) to obtain \(u'\). The derivative is calculated as follows:\(u' = \frac{d}{dx}(x^2 - 2x + 2) = 2x - 2\).

Step 4: Differentiate the Second Function Component

For \(v = e^x\), the derivative is straightforward since the derivative of \(e^x\) with respect to \(x\) is itself: \(v' = \frac{d}{dx}(e^x) = e^x\).

Step 5: Compile the Product Rule Expression

Substitute \(u\), \(u'\), \(v\), and \(v'\) into the product rule formula:\[ \frac{dy}{dx} = (2x - 2)e^x + (x^2 - 2x + 2)e^x \]

Step 6: Simplify the Expression

Combine the terms in the expression:\[ \frac{dy}{dx} = (2x - 2 + x^2 - 2x + 2)e^x \]Simplify within the parentheses, noting that \(2x\) cancels \(-2x\) and \(-2\) cancels \(2\), leaving:\[ \frac{dy}{dx} = x^2 e^x \]

Step 7: Write the Final Derivative

The simplified derivative of the function is:\( \frac{dy}{dx} = x^2 e^x \). This is the final expression for the derivative.

Key Concepts

Product Rule

When differentiating a product of two functions in calculus, the product rule is essential. The rule is pretty straightforward: if you have two functions, say \( u \) and \( v \), their product \( y = uv \) can be differentiated using the product rule formula: \( \frac{dy}{dx} = u'v + uv' \). This formula shows us that to find the derivative of the product, we need to:

• Differentiate the first function \( u \) while keeping \( v \) unchanged.
• Then differentiate the second function \( v \) while keeping \( u \) unchanged.
• Finally, add these two results together.

In the given exercise, we used the product rule on the function \( y = (x^2 - 2x + 2)e^x \) by identifying \( u = x^2 - 2x + 2 \) and \( v = e^x \). By following the steps of the product rule, we were able to find the derivative effectively. This rule is fundamental in calculus because it extends the basic concept of differentiation to more complex operations involving multiplication of functions.

Differentiation

Differentiation is a process in calculus used to find the rate at which a function is changing at any given point. It's like asking how fast something is going right now, rather than finding the average speed over time.
In the context of the exercise, differentiation helps us understand how the function \( y = (x^2 - 2x + 2)e^x \) changes with respect to \( x \). Differentiation involves finding derivatives, which can be thought of as the "slopes" of curves at given points. These derivatives can tell us:

• If the function is increasing or decreasing.
• Detecting peaks and valleys in the curve (maximum and minimum points).
• The concavity of the curve (how it "bends").

In our solution, differentiation required us to explore the individual derivatives of the components \( u \) and \( v \) separately, and then apply the product rule to get the complete derivative. Understanding and mastering differentiation rules is a stepping stone to tackling more complicated calculus problems.

Calculus

Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. It's essentially about change and motion, which makes it incredibly powerful for modelling and solving problems involving continuously changing quantities.
In our specific example, we looked at a task from differential calculus, a sub-field that primarily deals with how functions change. The goal was to find how \( y \), a function dependent on \( x \), changes with \( x \) by finding its derivative.

• Calculus helps us predict future behavior of dynamic systems by understanding current trends.
• It provides tools to maximize or minimize functions, which is key in optimisation problems.
• It allows us to calculate areas and volumes of complex shapes, going beyond basic geometry.

For students, grasping calculus concepts like differentiation and the product rule is crucial. They build the foundation for advanced studies in science and engineering, where calculus is used to design and optimize systems from electrical circuits to motion dynamics in physics.