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Chapter 3: Problem 5

Write derivative formulas for the functions. $$ f(x)=\left(12.8 x^{2}+3.7 x+1.2\right)\left[29\left(1.7^{x}\right)\right] $$

Short Answer

Expert verified
Use product rule: derivative is \( 29 \times 1.7^x \left( (25.6x + 3.7) + (12.8x^2 + 3.7x + 1.2) \ln(1.7) \right) \).

Step by step solution

01

Identify the Function Type

The given function is a product of two functions. Let's denote \( u(x) = 12.8x^2 + 3.7x + 1.2 \) and \( v(x) = 29 \times 1.7^x \). Therefore, we need to apply the product rule to find the derivative \( f'(x) \).
02

State the Product Rule

The product rule states that if \( u(x) \) and \( v(x) \) are two differentiable functions, then the derivative of their product is given by: \[ (u(x)v(x))' = u'(x)v(x) + u(x)v'(x) \]
03

Differentiate \( u(x) \)

The function \( u(x) = 12.8x^2 + 3.7x + 1.2 \) is a polynomial. The derivative is found using the power rule: \[ u'(x) = 2 \times 12.8 \times x^{2-1} + 3.7 = 25.6x + 3.7 \]
04

Differentiate \( v(x) \)

The function \( v(x) = 29 \times 1.7^x \) is an exponential function. The derivative of \( a^x \) (where \( a \) is a constant) is \( a^x \ln(a) \). Therefore: \[ v'(x) = 29 \times 1.7^x \ln(1.7) \]
05

Apply the Product Rule

Substitute \( u'(x) \, , \, u(x) \, , \, v'(x) \, \text{and} \, v(x) \) into the product rule formula: \[ f'(x) = (25.6x + 3.7)(29 \times 1.7^x) + (12.8x^2 + 3.7x + 1.2)(29 \times 1.7^x \ln(1.7)) \]
06

Simplify the Derivative Expression

Factor out \( 29 \times 1.7^x \) if needed, to simplify:\[ f'(x) = 29 \times 1.7^x \left( (25.6x + 3.7) + (12.8x^2 + 3.7x + 1.2) \ln(1.7) \right) \]

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Key Concepts

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

Product Rule
The product rule is a fundamental tool in calculus used to find the derivative of products of two functions. This rule is essential when you are dealing with a function defined as the product of two differentiable functions.
This rule states:
  • If you have two functions, good candidates are named as \( u(x) \) and \( v(x) \), the derivative of their product \( u(x)v(x) \) is given by:
\[ (u(x)v(x))' = u'(x)v(x) + u(x)v'(x) \]In simple terms, you'll differentiate each function separately, multiply each derivative by the other original function, and then sum the two.
This approach is particularly useful when the direct application of regular derivative rules becomes less straightforward due to the multiplication involved.
Polynomial Function
A polynomial function is a mathematical expression involving a sum of powers in one or more variables multiplied by coefficients.
In our example, \( u(x) = 12.8x^2 + 3.7x + 1.2 \) is a polynomial function.
  • Polynomial functions are relatively straightforward to deal with, especially when finding derivatives, thanks to the power rule.
  • These functions can include terms like linear terms (\( x \)), quadratic terms (\( x^2 \)), cubic terms (\( x^3 \)), and so on.
They're smooth and continuous, which means they don't have any gaps or sharp turns. This makes them ideal for modeling various real-world scenarios.
Exponential Function
Exponential functions involve equations where the variable appears in an exponent.
In our derivative exercise, \( v(x) = 29 \times 1.7^x \) is treated as an exponential function.
  • An exponential function with base \( a \), such as \( a^x \), grows or decays at a constant rate, which is determined by the base value.
  • These functions are characterized by their rapid increase (or decrease) as \( x \) changes.
To find their derivatives, a special rule applies: \( (a^x)' = a^x \ln(a) \). The ln term (natural logarithm) accounts for the exponential nature, which makes these derivatives non-linear and interesting.
Power Rule
The power rule is a basic derivative rule used for finding the derivatives of polynomial terms. This rule is a straightforward method applied to functions where variables are raised to a power.
If you have a term in the form of \( ax^n \), the power rule tells you:
  • The derivative is \( n \cdot ax^{n-1} \).
For instance, applying the power rule to \( 12.8x^2 \), we get the derivative as \( 2 \times 12.8x^{2-1} = 25.6x \).
This method helps quickly determine the slope of any polynomial function at any given point, which is why it's a cornerstone of differentiation.

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Most popular questions from this chapter

Evaluate the limit. If the limit is of an indeterminate form, indicate the form and use L'H么pital's Rule to evaluate the limit. $$ \lim _{x \rightarrow 5} \frac{(x-1)^{0.5}-2}{x^{2}-25} $$

Demand Let \(D(x)\) be the demand (in units) for a new product when the price is \(x\) dollars. a. Write sentences interpreting the following: i. \(D(6.25)=1000\) ii. \(D^{\prime}(6.25)=-50\) b. Write a formula for the revenue \(R(x)\) generated from the sale of the product when the price is \(x\) dollars. Assume demand is equal to the number sold. c. Calculate and interpret \(R^{\prime}(x)\) when \(x=6.25\).

Paint Sales The sales of Sherwin-Williams paint in different regional markets depends on several input variables. One variable that partially drives selling price, which in turn affects sales, is the cost to get the product to market. When other input variables are held constant, sales can be modeled as $$ s(x)=597.3\left(0.921^{4 x+12}\right) \text { thousand gallons } $$ when it costs \(x\) dollars to get a gallon of paint to market, data from \(0 \leq x \leq 2\) (Source: Based on data from Sherwin-Williams as reported in J. McGuigan et al., Managerial Economics, Stamford, CT: Cengage, 2008\()\) a. How many gallons of paint are sold when it costs 75 cents for a gallon to reach the market? b. How quickly is sales changing when \(x=0.75 ?\)

Identify the indeterminate form of each limit. Use L'H么pital's Rule to evaluate the limit of any indeterminate forms. $$ \lim _{t \rightarrow \infty} \frac{e^{2 t}}{2 t} $$

Write derivative formulas for the functions. $$ f(x)=\left(3 x^{2}+15 x+7\right)\left(32 x^{3}+49\right) $$
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