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Magazine Chapter 3: Problem 10
\(3-32\) Differentiate the function. \(h(x)=(x-2)(2 x+3)\)
Short Answer
Expert verified
The derivative is \(h'(x) = 4x - 1\).
Step by step solution
01
Identify the Product Rule
The given function is a product of two functions, i.e., \(h(x)=(x-2)(2x+3)\). Therefore, we need to apply the product rule to differentiate. The product rule states that if \(u(x)\) and \(v(x)\) are functions of \(x\), then \((uv)' = u'v + uv'\). Identify \(u(x) = x-2\) and \(v(x) = 2x+3\).
02
Differentiate Each Function
Differentiate each part of the product. For \(u(x)=x-2\), the derivative is \(u'(x)=1\). For \(v(x) = 2x+3\), the derivative is \(v'(x) = 2\). This is because the derivative of \(x\) is \(1\) and the derivative of a constant is \(0\).
03
Apply the Product Rule
Now, use the product rule: \(h'(x) = u'(x)v(x) + u(x)v'(x)\). Substitute the derivatives and original functions: \(h'(x) = (1)(2x+3) + (x-2)(2)\).
04
Simplify the Expression
Simplify the expression obtained from applying the product rule: \(h'(x) = 2x+3 + 2(x-2)\). Distribute the second term: \(h'(x) = 2x+3 + 2x - 4\). Combine like terms: \(h'(x) = 4x - 1\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivative of a Product
Breaking down the concept of a derivative of a product in calculus is key to understanding how to differentiate certain functions. In the given function, **h(x) = (x-2)(2x+3)**, you notice it is the product of two smaller parts, **u(x) = x - 2** and **v(x) = 2x + 3**.
The idea behind the product rule, which is **\((uv)' = u'v + uv'\)**, is to deal with the challenge of differentiating two functions multiplied together. This method allows you to manage each function's derivative independently before combining them. Identifying these sub-functions in the product is crucial for setting up the differentiation process.
The idea behind the product rule, which is **\((uv)' = u'v + uv'\)**, is to deal with the challenge of differentiating two functions multiplied together. This method allows you to manage each function's derivative independently before combining them. Identifying these sub-functions in the product is crucial for setting up the differentiation process.
- The derivative measures how a function changes as its input changes.
- Using the product rule helps manage functions that are products of other simpler functions.
- Learning to recognize the structure of a product within an expression is the first step to successful differentiation.
Differentiation Steps
Step-by-step differentiation is fundamental in tackling calculus problems accurately. Here鈥檚 how you navigate through the process for our example function.
Once you've identified **u(x) = x - 2** and **v(x) = 2x + 3**, the next phase is differentiation. The derivative of **u(x)**, which is **u'(x)**, comes out to **1**, because the derivative of **x** is **1**, and constants have derivatives equal to zero. Similarly, for **v(x)**, the derivative **v'(x) = 2**, as you're again applying basic differentiation rules to a linear term and a constant.
Once you've identified **u(x) = x - 2** and **v(x) = 2x + 3**, the next phase is differentiation. The derivative of **u(x)**, which is **u'(x)**, comes out to **1**, because the derivative of **x** is **1**, and constants have derivatives equal to zero. Similarly, for **v(x)**, the derivative **v'(x) = 2**, as you're again applying basic differentiation rules to a linear term and a constant.
- Establishing derivatives for each function part independently keeps calculations straightforward.
- Always verify differentiation rules, such as constants having a derivative of zero, are applied correctly.
Simplifying Derivatives
After applying the product rule, it's time for simplification, ensuring the result is as clear and compact as possible. In the example, you reached a derivative expression,
**h'(x) = (1)(2x+3) + (x-2)(2)**, by plugging the derivatives and originals back into the product rule formula.
This equation simplifies by expanding and collecting like terms: **h'(x) = 2x+3 + 2x - 4**, bringing together terms yields **h'(x) = 4x - 1**.
This equation simplifies by expanding and collecting like terms: **h'(x) = 2x+3 + 2x - 4**, bringing together terms yields **h'(x) = 4x - 1**.
- Simplifying derivatives reduces potential errors and makes functions much easier to work with.
- Combining like terms and distributing correctly helps result in a concise expression.
Calculus Problem-Solving
Problem-solving with calculus involves a strategic approach to breaking down functions and working through differentiation. The process isn鈥檛 just about applying rules but understanding why and when to use them. In our function, recognizing it as a product dictates the method, while clearly defining steps such as identifying parts, differentiating accurately, applying the rules, then simplifying is a cycle you will replicate across various problems.
- Effective calculus problem-solving combines methodical steps with creative logical reasoning.
- Translating from theory to practice is essential to cement understanding, making practice not just mechanical, but a way to see patterns.
- The skills you develop here are transferable to many mathematical contexts, showcasing the power and flexibility of calculus techniques.
