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$$\text { Find } f^{\prime}(x)$$. $$f(x)=(x-2)\left(x^{2}+2 x+4\right)$$

Short Answer

Expert verified
\( f'(x) = 3x^2 - 4 \)

Step by step solution

01

Identify the Derivative Rule Needed

To find the derivative of the function \( f(x) = (x-2)\left(x^2 + 2x + 4\right) \), we recognize that it is a product of two functions. Therefore, we will use the **Product Rule** for differentiation.
02

Break Down the Functions

In the Product Rule, if \( u(x) = x-2 \) and \( v(x) = x^2 + 2x + 4 \), the rule states that the derivative \( f'(x) \) is \( u'(x)v(x) + u(x)v'(x) \). We will differentiate these two functions separately.
03

Differentiate Each Function

First, find the derivative of \( u(x) \):\[u'(x) = (x-2)' = 1 \]Next, find the derivative of \( v(x) \):\[v'(x) = (x^2 + 2x + 4)' = 2x + 2 \]
04

Apply the Product Rule

Substitute the derivatives into the Product Rule formula:\[f'(x) = u'(x)v(x) + u(x)v'(x) \f'(x) = (1)(x^2 + 2x + 4) + (x-2)(2x + 2) \]
05

Simplify the Expression

First, distribute in each part of the expression:\[f'(x) = x^2 + 2x + 4 + 2x^2 + 2x - 4x - 4 \]Combine like terms:\[f'(x) = 3x^2 + 0x - 4 \f'(x) = 3x^2 - 4 \]
06

Final Result

Thus, the derivative is:\( f'(x) = 3x^2 - 4 \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differentiation
Differentiation is a fundamental concept in calculus. It involves finding the rate at which a function changes at any given point. Unlike basic arithmetic operations such as addition and subtraction, differentiation gives us an instantaneous rate of change. In simpler terms, it's like calculating how fast something is moving at a specific moment rather than over an interval.
To differentiate a function, we apply rules called 'differentiation rules'. These rules help us find the derivative, which is essentially the slope of the tangent line to the curve of the function at a particular point. Differentiation can be used on different types of functions including polynomials, exponentials, and trigonometric functions. Understanding differentiation is crucial for solving problems in physics, engineering, and other scientific fields, where predicting the behavior of changing quantities is necessary. It forms the first step towards understanding more complex topics in calculus.
Derivative
A derivative represents how a function's output value changes as its input value changes. It is a crucial building block of calculus, providing insights into the behavior of functions. When we talk about the derivative of a function, we're considering the function's rate of change at each point.
Mathematically, if we have a function \( f(x) \), the derivative is denoted as \( f'(x) \) or \( \frac{df}{dx} \). For instance, if \( f(x) = x^2 \), its derivative \( f'(x) = 2x \). This means that for a very small change in \( x \), the function's output changes approximately by \( 2x \).
  • Higher derivatives provide additional insights, such as concavity or point of inflection of the function.
  • The first derivative gives the slope of the tangent line.
By understanding derivatives, you can analyze and predict the behavior of graphs of functions. They are extensively used in every branch of science for optimization and modeling.
Product of Functions
When dealing with the product of two functions in calculus, a specific rule called the Product Rule is applied during differentiation. This rule is essential when you have two functions multiplying each other and you need to find the derivative of this product.
In our exercise, we have a function \( f(x) =(x-2)(x^2+2x+4) \). Here, we can identify two distinct functions: \( u(x) = x-2 \) and \( v(x) = x^2+2x+4 \). The Product Rule states that the derivative of a product of two functions is given by:\[ f'(x) = u'(x)v(x) + u(x)v'(x) \]This formula indicates that we must differentiate each function separately and then add the product of derivative pairs.
  • First, differentiate \( u(x) \) and \( v(x) \) individually.
  • Next, use the derivatives in the Product Rule formula.
  • Finally, simplify the expression to find the overall derivative of the function.
The Product Rule is vital for handling complex functions composed of multiple smaller functions. Knowing it allows you to dissect the function into manageable parts and solve them systematically. Before tackling such problems, ensure you can comfortably differentiate simpler functions.

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Most popular questions from this chapter

A paint manufacturing company estimates that it can sell \(g=f(p)\) gallons of paint at a price of \(p\) dollars per gallon. (a) What are the units of \(d g / d p ?\) (b) In practical terms, what does \(d g / d p\) mean in this case? (c) What can you say about the sign of \(d g / d p ?\) (d) Given that \(d g /\left.d p\right|_{p=10}=-100,\) what can you say about the effect of increasing the price from \(\$ 10\) per gallon to \(\$ 11\) per gallon?

You are asked in these exercises to determine whether a piecewise-defined function \(f\) is differentiable at a value \(x=x_{0}\) where \(f\) is defined by different formulas on different sides of \(x_{0}\). You may use without proof the following result, which is a consequence of the Mean-Value Theorem (discussed in Section 4.8 ). Theorem. Let \(f\) be continuous at \(x_{0}\) and suppose that \(\lim _{x \rightarrow x_{0}} f^{\prime}(x)\) exists. Then \(f\) is differentiable at \(x_{0}\) and \(f^{\prime}\left(x_{0}\right)=\lim _{x \rightarrow \infty} f^{\prime}(x)\) Let $$ f(x)=\left\\{\begin{array}{ll} x^{2}-16 x, & x<9 \\ \sqrt{x}, & x \geq 9 \end{array}\right. $$ Is \(f\) continuous at \(x=9 ?\) Determine whether \(f\) is differentiable at \(x=9 .\) If so, find the value of the derivative there.

$$ \begin{array}{l} \text { Let } f(x)=x^{8}-2 x+3 ; \text { find } \\ \qquad \lim _{w \rightarrow 2} \frac{f^{\prime}(w)-f^{\prime}(2)}{w-2} \end{array} $$

Writing Write a paragraph that explains what it means for a function to be differentiable. Include examples of functions that are not differentiable as well as examples of functions that are differentiable.

Given the following table of values, find the indicated derivatives in parts (a) and (b). $$\begin{array}{|c|c|c|} \hline x & f(x) & f^{\prime}(x) \\ \hline 2 & 1 & 7 \\\ \hline 8 & 5 & -3 \\ \hline \end{array}$$ (a) \(g^{\prime}(2),\) where \(g(x)=[f(x)]^{3}\) (b) \(h^{\prime}(2),\) where \(h(x)=f\left(x^{3}\right)\)

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