Problem 37
Question
Find the derivatives of the functions in Exercises \(23-50\). $$y=\left(x^{2}-2 x+2\right) e^{5 x / 2}$$
Step-by-Step Solution
Verified Answer
The derivative is \( y' = e^{5x/2} (\frac{5}{2}x^2 - 3x + 3) \).
1Step 1: Understanding the Product Rule
To find the derivative of the function \( y = (x^2 - 2x + 2) e^{5x/2} \), notice that it is a product of two functions: \( u(x) = x^2 - 2x + 2 \) and \( v(x) = e^{5x/2} \). We will use the product rule, which states \( (uv)' = u'v + uv' \).
2Step 2: Finding the Derivative of the First Function
Calculate the derivative of \( u(x) = x^2 - 2x + 2 \). The derivative, \( u'(x) \), is the derivative of each term separately: \( u'(x) = 2x - 2 \).
3Step 3: Finding the Derivative of the Second Function
Calculate the derivative of \( v(x) = e^{5x/2} \). Use the chain rule where \( v'(x) = (e^{5x/2})' = e^{5x/2} \cdot \frac{5}{2} \). Thus, \( v'(x) = \frac{5}{2} e^{5x/2} \).
4Step 4: Applying the Product Rule
Substitute \( u(x), u'(x), v(x), \) and \( v'(x) \) into the product rule formula: \( y' = (2x - 2)e^{5x/2} + (x^2 - 2x + 2) \cdot \frac{5}{2}e^{5x/2} \).
5Step 5: Simplifying the Expression
Simplify the expression for \( y' \): \( y' = e^{5x/2} \left[ (2x - 2) + \frac{5}{2}(x^2 - 2x + 2) \right] \). Simplify inside the brackets: \( (2x - 2) = 2x - 2 \) and \( \frac{5}{2}(x^2 - 2x + 2) = \frac{5}{2}x^2 - 5x + 5 \). Combine terms to get: \( y' = e^{5x/2} \left( \frac{5}{2}x^2 - 3x + 3 \right) \).
Key Concepts
product rulechain rulesimplifying expressions
product rule
The product rule is essential when dealing with derivatives of functions that are multiplied together.
When a function is a product of two distinct functions, such as \( u(x) \) and \( v(x) \), the product rule offers a formula to find its derivative.
It states that the derivative of a product \((uv)'\) is \(u'v + uv'\). This means you take the derivative of one function and multiply it by the other, then add the result of the opposite. For example, in the expression \( y = (x^2 - 2x + 2) e^{5x/2} \), we identify:
This formula captures how both functions and their derivatives contribute to the rate of change of the product as a whole.
When a function is a product of two distinct functions, such as \( u(x) \) and \( v(x) \), the product rule offers a formula to find its derivative.
It states that the derivative of a product \((uv)'\) is \(u'v + uv'\). This means you take the derivative of one function and multiply it by the other, then add the result of the opposite. For example, in the expression \( y = (x^2 - 2x + 2) e^{5x/2} \), we identify:
- \( u(x) = x^2 - 2x + 2 \)
- \( v(x) = e^{5x/2} \)
- \( u'(x) = 2x - 2 \)
- \( v'(x) = \frac{5}{2} e^{5x/2} \)
This formula captures how both functions and their derivatives contribute to the rate of change of the product as a whole.
chain rule
The chain rule comes into play when we need to differentiate a composite function.
This rule is vital when a function is inside another function, often encountered when dealing with exponential functions. The chain rule states that if you have a function \( g(x) \) inside another function \( f(g(x)) \), the derivative is \( f'(g(x)) \cdot g'(x) \). In our exercise, we need the chain rule for \( v(x) = e^{5x/2} \). The outer function \( f(u) = e^u \) and the inner function \( g(x) = 5x/2 \) together form \( f(g(x)) = e^{5x/2} \).
The derivative of \( e^{5x/2} \) uses:
The chain rule simplifies finding derivatives of such composite expressions and is a powerful tool for calculus tasks.
This rule is vital when a function is inside another function, often encountered when dealing with exponential functions. The chain rule states that if you have a function \( g(x) \) inside another function \( f(g(x)) \), the derivative is \( f'(g(x)) \cdot g'(x) \). In our exercise, we need the chain rule for \( v(x) = e^{5x/2} \). The outer function \( f(u) = e^u \) and the inner function \( g(x) = 5x/2 \) together form \( f(g(x)) = e^{5x/2} \).
The derivative of \( e^{5x/2} \) uses:
- Outer derivative \( f'(u) = e^u \)
- Inner derivative \( g'(x) = \frac{5}{2} \)
The chain rule simplifies finding derivatives of such composite expressions and is a powerful tool for calculus tasks.
simplifying expressions
Simplifying expressions is a critical final step in solving derivative problems.
This process helps in presenting the answer in its simplest form, making it easier to interpret and use in subsequent analyses or calculations.After applying the derivative rules, the expression for \( y' \) may initially seem complex.
The expression \( y' = e^{5x/2} \left[ (2x - 2) + \frac{5}{2}(x^2 - 2x + 2) \right] \) can be simplified by:
Proper simplification ensures clarity and correctness in mathematical solutions.
This process helps in presenting the answer in its simplest form, making it easier to interpret and use in subsequent analyses or calculations.After applying the derivative rules, the expression for \( y' \) may initially seem complex.
The expression \( y' = e^{5x/2} \left[ (2x - 2) + \frac{5}{2}(x^2 - 2x + 2) \right] \) can be simplified by:
- Distributing constants and simplifying products step-by-step.
- Combining like terms inside the brackets.
- \( (2x - 2) = 2x - 2 \)
- \( \frac{5}{2}(x^2 - 2x + 2) = \frac{5}{2}x^2 - 5x + 5 \)
Proper simplification ensures clarity and correctness in mathematical solutions.
Other exercises in this chapter
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