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In Exercises 1 through \(34,\) find the derivative. $$ f(x)=\left(x^{2}+5\right)\left(x^{5}-3 x+2\right) $$

Short Answer

Expert verified
The derivative is \( f'(x) = 7x^6 + 25x^4 - 9x^2 + 4x - 15 \).

Step by step solution

01

Identify the Expression Type

The function given is a product of two polynomials, \( f(x) = (x^2 + 5)(x^5 - 3x + 2) \). In order to find its derivative, we will use the product rule.
02

Recall the Product Rule

The product rule states that if you have a function \( h(x) = u(x) \cdot v(x) \), then its derivative is \( h'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) \). Here, \( u(x) = x^2 + 5 \) and \( v(x) = x^5 - 3x + 2 \).
03

Differentiate the First Function

Differentiate \( u(x) = x^2 + 5 \). The derivative, \( u'(x) \), is found by applying the power rule: \( u'(x) = 2x \).
04

Differentiate the Second Function

Differentiate \( v(x) = x^5 - 3x + 2 \). Applying the power rule term by term gives \( v'(x) = 5x^4 - 3 \).
05

Apply the Product Rule

Using the product rule: \[ f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) = (2x)(x^5 - 3x + 2) + (x^2 + 5)(5x^4 - 3) \].
06

Simplify the Expression

Expand each term: - \( 2x(x^5 - 3x + 2) = 2x^6 - 6x^2 + 4x \)- \( (x^2 + 5)(5x^4 - 3) = 5x^6 + 25x^4 - 3x^2 - 15 \) Combine the like terms to get the derivative: \[ f'(x) = 7x^6 + 25x^4 - 9x^2 + 4x - 15 \]

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Key Concepts

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

Product Rule
When you have a function that is a product of two other functions, the product rule is your best friend in calculus. It's a technique used to differentiate such product functions, and it states that if you have two functions, say \( u(x) \) and \( v(x) \), then the derivative of the product \( h(x) = u(x) \cdot v(x) \) is given by:
  • \( h'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) \)
This formula combines the derivatives of the two functions with their original forms. It makes sure we account for the rate of change in both parts of the product. So, if you're looking at \( f(x) = (x^2 + 5)(x^5 - 3x + 2) \), using this rule splits the differentiation task into smaller, manageable parts.
Getting comfortable with the product rule is essential as it frequently appears in calculus when dealing with polynomial functions.
Power Rule
The power rule is another fundamental tool you'll use a lot in calculus, especially when differentiating polynomials. This rule is all about simplifying the process of finding derivatives for terms of the form \( x^n \). For such a term, the power rule tells us:
  • The derivative is \( nx^{n-1} \).
Let's break this down with an example. If you have \( x^2 \), the derivative using the power rule would be \( 2x^{2-1} = 2x \). For a term like \( x^5 \), it becomes \( 5x^{5-1} = 5x^4 \).
When you apply the power rule to each term in a polynomial, it makes differentiation straightforward and mechanical. This simplicity is very helpful in dealing with polynomials, like when finding the derivative of \( u(x) = x^2 + 5 \) and \( v(x) = x^5 - 3x + 2 \) as part of a product rule application.
Polynomials
Polynomials are algebraic expressions that consist of several terms, each being a product of a constant multiplier and one or more variables raised to a whole number exponent. They form the building blocks for many functions in mathematics.
  • Examples include: \( x^2 + 5 \) and \( x^5 - 3x + 2 \).
  • They are often written in standard form, from highest degree to lowest.
One of the key properties of polynomials is that they are smooth and continuous, which is why they are so prevalent in calculus.
When differentiating polynomials, each term can be handled separately by applying the power rule. This means differentiating a polynomial might feel like handling a series of straightforward calculations.
The final derivative is simple to construct by recomposing the differentiated terms, just as we did when finding \( f'(x) \) for the polynomial product.

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In Exercises 1 through \(34,\) find the derivative. $$ \left(x^{2}+x+2\right) e^{x} $$

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