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In Exercises 1 through \(34,\) find the derivative. $$ f(x)=\left(x^{2}+3\right)\left(x^{4}+x+1\right)^{5} $$

Short Answer

Expert verified
The derivative is \( f'(x) = 2x (x^4 + x + 1)^5 + 5(x^2 + 3)(x^4 + x + 1)^4 (4x^3 + 1). \)

Step by step solution

01

Identify the Function Type

The function given is a product of two functions: \( u(x) = x^2 + 3 \) and \( v(x) = (x^4 + x + 1)^5 \). To find the derivative of the product, we need to use the product rule.
02

Apply the Product Rule

The product rule states that the derivative of a product of two functions \( u(x) \) and \( v(x) \) is given by \[(u \, v)' = u' \, v + u \, v'.\]Next, we need to find \( u'(x) \) and \( v'(x) \).
03

Find \( u'(x) \)

The function \( u(x) = x^2 + 3 \) is a simple polynomial. Its derivative is \[u'(x) = 2x.\]
04

Find \( v'(x) \) Using the Chain Rule

The function \( v(x) = (x^4 + x + 1)^5 \) requires the chain rule to differentiate. The outer function is \( g(y) = y^5 \) with \( y = x^4 + x + 1 \). The derivative of \( g(y) \) with respect to \( y \) is \( 5y^4 \), and the derivative of \( y \) with respect to \( x \) is \( 4x^3 + 1 \).Thus, \[v'(x) = 5(x^4 + x + 1)^4 \, (4x^3 + 1).\]
05

Substitute into the Product Rule

Now substitute \( u(x) \), \( u'(x) \), \( v(x) \), and \( v'(x) \) into the product rule formula:\[f'(x) = (2x)(x^4 + x + 1)^5 + (x^2 + 3)(5(x^4 + x + 1)^4 (4x^3 + 1)).\]
06

Simplify the Expression

Simplify the derivative by combining like terms or expanding as needed, though it's often fine to keep it in its factored form for simplicity. The derivative \( f'(x) \) is:\[f'(x) = 2x (x^4 + x + 1)^5 + 5(x^2 + 3)(x^4 + x + 1)^4 (4x^3 + 1).\]

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Key Concepts

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

The Product Rule
The product rule is a fundamental technique in calculus used to differentiate functions that are products of separate functions. When you have a function like \( f(x) = u(x) \times v(x) \), where both \( u \) and \( v \) are functions of \( x \), the product rule comes to the rescue. It allows us to find the derivative of the entire product without first expanding it, which can be a lifesaver for complex expressions.

A straightforward way to remember the product rule is by the formula:
  • \((u \, v)' = u' \, v + u \, v'\)
This formula states that to find the derivative of \( u(x) \times v(x) \), we take the derivative of \( u \), multiply it by \( v \), then add \( u \) times the derivative of \( v \).

This method is particularly useful because it simplifies the process of differentiation when each component of the product is not easily broken down or simplified further. The product rule allows for strategic differentiation without unnecessary expansion.
The Chain Rule
The chain rule is another essential tool in calculus, used when dealing with composite functions. These functions are essentially functions within functions, like \( v(x) = (x^4 + x + 1)^5 \) in this exercise. This is where the chain rule shines.

For a composite function \( f(g(x)) \), the chain rule states that the derivative \( f'(x) \) is
  • \( f'(x) = f'(g(x)) \times g'(x) \)
Here's a simple way to think about it:
  • First, take the derivative of the outer function \( f \) with respect to \( g \).
  • Then multiply it by the derivative of the inner function \( g \) with respect to \( x \).
In the given problem, the outer function is \( y^5 \) and the inner function is \( x^4 + x + 1 \).

By applying the chain rule, you efficiently differentiate these nested functions without unravelling the entire composite. It dramatically simplifies finding derivatives, especially in situations with powers and nested expressions.
Polynomial Differentiation
Polynomial differentiation is the process of finding the derivative of polynomial functions. Polynomials consist of terms like \( x^2 + 3 \), which are straightforward to handle because their derivatives can be determined using simple rules.

For any polynomial term of the form \( ax^n \), the power rule is applied:
  • The derivative is \( n \times ax^{n-1} \)
This rule simplifies the differentiation of polynomials by providing a clear pattern to follow.

Taking the derivative of \( x^2 + 3 \), you apply the power rule to each term separately. The constant term \(+3\) becomes \(0\) because constants have no rate of change with respect to \( x \).

Polynomials, due to their simplicity and familiarity, are excellent practice for learning differentiation basics. When these polynomials are parts of larger or more complex functions, which may require the product or chain rules, understanding their fundamental differentiation will greatly ease the process.

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Most popular questions from this chapter

Finance Fry \({ }^{15}\) studied 85 developing countries and found that the average percentage growth rate \(y\) in each country was approximated by the equation \(y=g(r)=\) \(-0.033 r^{2}+0.008 r^{3},\) where \(r\) is the real interest rate in the country. Find \(g^{\prime}(r)\), and explain what this term means. Find \(g^{\prime}(-2)\) and \(g^{\prime}(2) .\) Explain in each case what these two numbers mean. What is the significance of the sign of each number?

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Cost A large corporation already has cornered \(20 \%\) of the disposable diaper market and now wishes to undertake a very extensive advertising campaign to increase its share of this market. It is estimated that the advertising cost, \(C(x)\), in billions of dollars, of attaining an \(x \%\) share of this market is given by $$ C(x)=\frac{1}{10-0.1 x} $$ What is the marginal cost?

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