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In Activities 1 through \(30,\) for each of the composite functions, identify an inside function and an outside function and write the derivative with respect to \(x\) of the composite function. $$ f(x)=2^{\ln x} $$

Short Answer

Expert verified
The derivative of \( f(x) = 2^{\ln x} \) is \( \frac{2^{\ln x} \ln 2}{x} \).

Step by step solution

01

Determine the Inside Function

First, identify the inside function. For the given function \( f(x) = 2^{\ln x} \), the inside function is \( u(x) = \ln x \). This is because \( \ln x \) is inside the power of the base \( 2 \).
02

Identify the Outside Function

Next, identify the outside function which operates on the inside function \( u(x) = \ln x \). The outside function is \( g(u) = 2^u \), since \( 2^{u} \) is the operation applied to \( \ln x \).
03

Differentiate the Inside Function

Find the derivative of the inside function \( u(x) = \ln x \) with respect to \( x \). The derivative is \( \frac{du}{dx} = \frac{1}{x} \).
04

Differentiate the Outside Function with Respect to Its Argument

Differentiate the outside function \( g(u) = 2^u \) with respect to \( u \). We use the fact that the derivative of \( a^u \) with respect to \( u \) is \( a^u \ln(a) \). Therefore, \( \frac{dg}{du} = 2^u \ln 2 \).
05

Apply the Chain Rule

Combine the derivatives of the inside and outside functions using the chain rule. The derivative of the composite function \( f(x) = 2^{\ln x} \) is obtained by multiplying \( \frac{dg}{du} \) by \( \frac{du}{dx} \). Thus, \( \frac{d}{dx}[2^{\ln x}] = 2^{\ln x} \ln 2 \times \frac{1}{x} \).
06

Simplify the Derivative Expression

Combine and simplify the expression to get the final derivative: \[ \frac{d}{dx}[2^{\ln x}] = \frac{2^{\ln x} \ln 2}{x} \].

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Key Concepts

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

Chain Rule
The chain rule is a fundamental technique in calculus used to differentiate composite functions. Composite functions are those that apply multiple functions in sequence. In simpler terms, a function inside another function. This makes the derivative process slightly complex. The chain rule provides a structured method to tackle this.

For a composite function, if you have \(y = f(g(x))\), the chain rule states the derivative of y with respect to x is the product of the derivative of the outside function and the derivative of the inside function. Mathematically, it's expressed as \(\frac{dy}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx}\).

  • First, differentiate the outside function while keeping the inside function intact.
  • Then, multiply the result by the derivative of the inside function.
This process allows us to break down and simplify the otherwise complex differentiation process.
Inside Function
The inside function is a crucial component for utilizing the chain rule. It is the function that resides within another function. Identifying this correctly ensures the success of using the chain rule.

For example, in our exercise with \(f(x) = 2^{\ln x}\), the inside function is \(u(x) = \ln x\). When we differentiate the composite function, it is this inside function that we handle first.

The derivative of our inside function \(u(x) = \ln x\) is \(\frac{du}{dx} = \frac{1}{x}\). By singling out this inside function, it simplifies the process of differentiation.
Outside Function
The outside function is the function that encapsulates the inside function. It acts on the result computed from the inside function. Identifying it is vital for applying the chain rule effectively.

For our specific exercise \(f(x) = 2^{\ln x}\), where \(u(x) = \ln x\) is the inside function, the outside function becomes \(g(u) = 2^u\). In simpler words, it's the \(2^{}\) acting on the \(\ln x\).

To find its derivative, we differentiate the outside function with respect to its argument \(u\). This gives us \(\frac{dg}{du} = 2^{u} \ln 2\). This step contributes to forming our final derivative using the chain rule.
Logarithmic Function
Logarithmic functions involve the logarithm and are a key part of calculus, especially dealing with analytics and exponentials. They often appear as inside functions in composite functions.

In our exercise, the logarithmic function is \(\ln x\). Logarithms transform exponential relationships and are often pivotal in simplifying the differentiation process. The natural logarithm \(\ln\) has particularly simple derivatives, such as \(\frac{d}{dx}(\ln x) = \frac{1}{x}\).

  • Logarithmic differentiation can simplify certain derivations.
  • They serve as critical transitions from multiplication to addition in equations.
Understanding how to differentiate logarithmic functions helps to unravel complex relationships in calculus efficiently.

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Most popular questions from this chapter

College Tuition The tuition \(x\) years from now at a private four-year college is projected to be $$ t(x)=24,072 e^{0.056 x} \text { dollars. } $$ a. Write the rate-of-change formula for tuition. b. What is the rate of change in tuition four years from now?

In Activities 1 through \(30,\) for each of the composite functions, identify an inside function and an outside function and write the derivative with respect to \(x\) of the composite function. $$ f(x)=72 e^{0.6 x} $$

Production Cost A manufacturing company has found that it can stock no more than one week's worth of perishable raw material for its manufacturing process. When purchasing this material, the company receives a discount based on the size of the order. Company managers have modeled a cost as \(C(u)=196.3+44.5 \ln u\) dollars to produce \(u\) units per week. Each quarter, improvements are made to the automated machinery to help boost production. The company has kept a record of the average units per week that were produced during the past 16 quarters. These data are given in the table. Average Weekly Production Levels \begin{tabular}{|c|c|c|c|} \hline Quarter & Production (units) & Quarter & Production (units) \\ \hline 1 & 2000 & 9 & 3000 \\ \hline 2 & 2070 & 10 & 3200 \\ \hline 3 & 2160 & 11 & 3410 \\ \hline 4 & 2260 & 12 & 3620 \\ \hline 5 & 2380 & 13 & 3880 \\ \hline 6 & 2510 & 14 & 4130 \\ \hline 7 & 2660 & 15 & 4410 \\ \hline 8 & 2820 & 16 & 4690 \\ \hline \end{tabular} a. Find an appropriate model for production during the \(x\) th quarter. b. Use the company cost model along with the production model to write an expression modeling cost per week as a function of the quarter. c. Predict the company's cost per week in quarter 18 and quarter 20 d. Write an expression for the rate of change of the cost function in part \(b\). Based on the rate of change function, will cost ever decrease? Explain.

High School Dropouts (Historic) The table shows the number of students enrolled in the ninth through twelfth grades and the number of dropouts from those same grades in South Carolina for each school year from \(1980-1981\) through \(1989-1990\) $$ \begin{aligned} &\text { South Carolina High School Enrollment and Dropouts }\\\ &\begin{array}{|c|c|c|} \hline \text { School year } & \text { Enrollment } & \text { Dropouts } \\ \hline 1980-81 & 194,072 & 11,651 \\ \hline 1981-82 & 190,372 & 10,599 \\ \hline 1982-83 & 185,248 & 9314 \\ \hline 1983-84 & 182,661 & 9659 \\ \hline 1984-85 & 181,949 & 8605 \\ \hline 1985-86 & 182,787 & 8048 \\ \hline 1986-87 & 185,131 & 7466 \\ \hline 1987-88 & 183,930 & 7740 \\ \hline 1988-89 & 178,094 & 7466 \\ \hline 1989-90 & 172,372 & 5768 \\ \hline \end{array} \end{aligned} $$ a. Find cubic models for enrollment and the number of dropouts. Align both models to the number of years since \(1980-81\) b. Use the two models found in part \(a\) to construct an equation for the percentage of high school students who dropped out each year. c. Find the rate-of-change formula of the percentage of high school students who dropped out each year. d. Look at the rate of change for each school year from \(1980-81\) through \(1989-90 .\) In which school year was the rate of change smallest? When was it greatest?

Write derivative formulas for the functions. $$ f(x)=4 x \sqrt{3 x+2}+93 $$

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