Problem 13
Question
Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. \begin{equation}y=\left(x^{2}-2 x+2\right) e^{x}\end{equation}
Step-by-Step Solution
Verified Answer
The derivative is \( y' = (x^2 + 2x)e^x \).
1Step 1: Identify the Rule to Use
To find the derivative of the function \( y = (x^2 - 2x + 2) e^x \), we need to apply the Product Rule. This is because the function is a product of two functions of \( x \): \( u(x) = x^2 - 2x + 2 \) and \( v(x) = e^x \).
2Step 2: Apply the Product Rule
The Product Rule states that the derivative of a product \( u(x)v(x) \) is given by \( u'(x)v(x) + u(x)v'(x) \). We will find \( u'(x) \) and \( v'(x) \) and then substitute into the Product Rule formula.
3Step 3: Compute \( u'(x) \)
\( u(x) = x^2 - 2x + 2 \). Differentiate with respect to \( x \) to get \( u'(x) = 2x - 2 \).
4Step 4: Compute \( v'(x) \)
\( v(x) = e^x \). The derivative of \( e^x \) with respect to \( x \) is \( v'(x) = e^x \).
5Step 5: Substitute into Product Rule Formula
Using the Product Rule: \[ y' = u'(x)v(x) + u(x)v'(x) = (2x - 2)e^x + (x^2 - 2x + 2)e^x \]
6Step 6: Simplify the Expression
Combine the terms from the Product Rule application:\[ y' = (2x - 2)e^x + (x^2 - 2x + 2)e^x = (x^2 + 2x)e^x \]Factor out \( e^x \), the most common factor:\[ y' = e^x(x^2 + 2x) \].
Key Concepts
DifferentiationExponential FunctionsCalculus
Differentiation
Differentiation is the process used in calculus to calculate the rate at which a function changes at any point. It is a key concept that helps in finding the derivative, which tells how a function's output changes with respect to changes in its input. Differentiation is performed using specific rules that simplify the calculation of derivatives for various types of functions.
To differentiate a product of two functions, the Product Rule is employed. The Product Rule states that if you have two functions multiplied together, say \(u(x)\) and \(v(x)\), their derivative \(y'\) is given by:
This rule allows you to find how the combined effect of two varying quantities results in further changes.
With this problem, we have \(u(x) = x^2 - 2x + 2\) and \(v(x) = e^x\). Differentiating each function separately and applying the Product Rule, we accurately find the rate of change for the function \(y = (x^2 - 2x + 2) e^x\) with respect to \(x\).
To differentiate a product of two functions, the Product Rule is employed. The Product Rule states that if you have two functions multiplied together, say \(u(x)\) and \(v(x)\), their derivative \(y'\) is given by:
- \(u'(x)v(x) + u(x)v'(x)\)
This rule allows you to find how the combined effect of two varying quantities results in further changes.
With this problem, we have \(u(x) = x^2 - 2x + 2\) and \(v(x) = e^x\). Differentiating each function separately and applying the Product Rule, we accurately find the rate of change for the function \(y = (x^2 - 2x + 2) e^x\) with respect to \(x\).
Exponential Functions
Exponential functions, such as \(e^x\), play a fundamental role in differentiation and calculus. The base of the natural exponential function, \(e\), is approximately equal to 2.71828. It is unique because the rate of growth of \(e^x\) is proportional to its value.
When it comes to differentiation, the most remarkable property of \(e^x\) is that it is its own derivative:
In our context, the exponential function \(e^x\) remains unchanged in differentiation. No matter what point you evaluate \(e^x\) at, its slope or derivative remains \(e^x\), a trait that greatly simplifies calculus problems involving exponential growth or decay.
When it comes to differentiation, the most remarkable property of \(e^x\) is that it is its own derivative:
- If \(v(x) = e^x\), then \(v'(x) = e^x\)
In our context, the exponential function \(e^x\) remains unchanged in differentiation. No matter what point you evaluate \(e^x\) at, its slope or derivative remains \(e^x\), a trait that greatly simplifies calculus problems involving exponential growth or decay.
Calculus
Calculus is the branch of mathematics focusing on change. It offers tools to solve problems involving rates of change and areas under curves, among other applications. Calculus is divided into two main branches: differentiation and integration.
In this exercise, the focus is on differentiation—determining how a function changes at any given point. By using differentiation, especially rules like the Product Rule, complex expressions are broken down into manageable parts.
The power of calculus lies in translating physical problems into mathematical ones, solving them, and interpreting the results. It's the backbone of many scientific disciplines and essential for understanding changes and patterns in various contexts.
In this exercise, the focus is on differentiation—determining how a function changes at any given point. By using differentiation, especially rules like the Product Rule, complex expressions are broken down into manageable parts.
- First, identify functions and their roles in the expression.
- Use appropriate rules such as the Product Rule or Chain Rule as needed.
- Combine and simplify results to find the derivative.
The power of calculus lies in translating physical problems into mathematical ones, solving them, and interpreting the results. It's the backbone of many scientific disciplines and essential for understanding changes and patterns in various contexts.
Other exercises in this chapter
Problem 13
Solve the differential equations in Exercises \(9-22\) $$\frac{d y}{d x}=\sqrt{y} \cos ^{2} \sqrt{y}$$
View solution Problem 13
Use l'Hopital's rule to find the limits in Exercises \(7-50\) . $$ \lim _{t \rightarrow 0} \frac{\sin t^{2}}{t} $$
View solution Problem 14
Find the limits in Exercises \(13-20 .\) (If in doubt, look at the function's graph.) $$ \lim _{x \rightarrow-1^{+}} \cos ^{-1} x $$
View solution Problem 14
In Exercises \(13-24,\) find the derivative of \(y\) with respect to the appropriate variable. $$y=\frac{1}{2} \sinh (2 x+1)$$
View solution