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Chapter 4: Problem 6

Find the derivative of each function. $$ f(x)=\sqrt{\ln x} $$

Short Answer

Expert verified
The derivative is \( \frac{1}{2x\sqrt{\ln x}} \).

Step by step solution

01

Identify the Composition of Functions

The given function is \( f(x) = \sqrt{\ln x} \). Notice that it is a combination of two functions: an outer function \( g(u) = \sqrt{u} \) and an inner function \( u = \ln x \). This will require the chain rule for differentiation.
02

Differentiate the Outer Function

First, find the derivative of the outer function \( g(u) = \sqrt{u} \). The derivative is \( g'(u) = \frac{1}{2\sqrt{u}} \).
03

Differentiate the Inner Function

Now, differentiate the inner function \( u = \ln x \). The derivative is \( \frac{du}{dx} = \frac{1}{x} \).
04

Apply the Chain Rule

According to the chain rule, the derivative of \( f(x) = \sqrt{\ln x} \) is \( \frac{df}{dx} = g'(u) \cdot \frac{du}{dx} \). Substitute the derivatives from Step 2 and Step 3 into this formula: \[ \frac{df}{dx} = \frac{1}{2\sqrt{\ln x}} \cdot \frac{1}{x} = \frac{1}{2x\sqrt{\ln x}}. \]
05

Simplify the Derivative Expression

The result from Step 4 is already in a simplified form. The derivative of the function \( f(x) = \sqrt{\ln x} \) is \( \frac{1}{2x\sqrt{\ln x}} \).

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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 vital concept in calculus when dealing with derivatives of composite functions. Imagine you have a function that is made up of two or more functions. The Chain Rule helps you find the derivative of this type of function.
The general idea is simple: you take the derivative of the outermost function, then multiply it by the derivative of the inside function(s).
  • If you have a composed function like \(f(g(x))\), you would find the derivative by first differentiating the outer function, \(f'(g(x))\), then multiply by the inner function's derivative, \(g'(x)\).
In the original exercise, the Chain Rule guides us through differentiating each layer of the function \(f(x) = \sqrt{\ln x}\). This approach ensures you get the correct rate of change at every point on the function's curve.
Composition of Functions
Composition of functions is a fancy way of saying that one function is inside another. It's like placing a toy inside a box, and that box inside another box.
In terms of math, if you have \(g(u)\) and \(u = \ln x\), the composed function becomes \(g(\ln x)\).
  • The outer function is \(g(u) = \sqrt{u}\).
  • The inner function is \(u = \ln x\).
When you're asked to differentiate such a function, you鈥檙e essentially peeling back each layer, working your way from the outside in. Identifying this layering is crucial as it dictates how you apply the Chain Rule later during differentiation.
Differentiation
Differentiation is the core process in calculus for finding the rate at which one quantity changes relative to another.
Think of it as a way to find the slope of the tangent line at any point on a curve, which tells you how fast or slow something is changing.
  • When you differentiate an outer function like \(\sqrt{u}\), you get \(\frac{1}{2\sqrt{u}}\).
  • In the case of the inner function like \(\ln x\), the derivative is \(\frac{1}{x}\).
After differentiating both parts of a composed function, the Chain Rule combines these to give us the full derivative. In this original exercise, applying differentiation correctly took us from step-by-step details to the final derivative: \(\frac{1}{2x\sqrt{\ln x}}\). This informs us precisely how the original function behaves as \(x\) changes.

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Most popular questions from this chapter

The demand functions for distilled spirits and for beer are given below, where \(p\) is the retail price and \(D(p)\) is the demand in gallons per capita. For each demand function, find the elasticity of demand for any price \(p\). [Note: You will find, in each case, that demand is inelastic. This means that taxation, which acts like a price increase, is an ineffective way of discouraging liquor consumption, but is an effective way of raising revenue.] $$ D(p)=7.881 p^{-0.112} \quad \text { (for beer) } $$

If demand is inelastic and you raise prices, would you expect demand to increase or decrease, and strongly or weakly?

Use your graphing calculator to graph each function on a window that includes all relative extreme points and inflection points, and give the coordinates of these points (rounded to two decimal places). [Hint: Use NDERIV once or twice with ZERO.] (Answers may vary depending on the graphing window chosen.) $$ f(x)=e^{x}+e^{-x} $$

The demand function for automobiles in a dealership is given below, where \(p\) is the selling price. a. Use the method described in the Graphing Calculator Exploration on page 312 to find the elasticity of demand at a price of $$\$ 12,000$$. b. Should the dealer raise or lower the price from this level to increase revenue? c. Find the price at which elasticity equals 1. [Hint: Use INTERSECT. $$ D(p)=\frac{200}{8+e^{0.0001 p}} $$

A mug of beer chilled to 40 degrees, if left in a 70 -degree room, will warm to a temperature of \(T(t)=70-30 e^{-3.5 t}\) degrees in \(t\) hours. a. Find \(T(0.25)\) and \(T^{\prime}(0.25)\) and interpret your answers. b. Find \(T(1)\) and \(T^{\prime}(1)\) and interpret your answers.
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