Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 6: Problem 8
Evaluate the following derivatives. \(\frac{d}{d x}(\ln (\ln x))\)
Short Answer
Expert verified
Answer: The derivative of the function \(f(x) = \ln(\ln(x))\) is \(f'(x) = \frac{1}{x \ln(x)}\).
Step by step solution
01
Identify the outer and inner functions
In this exercise, the given function is \(\ln(\ln(x))\). The outer function is \(\ln(u)\), and the inner function is \(\ln(x)\). Let \(u = \ln(x)\).
02
Find the derivative of the outer function
We need to find the derivative of \(\ln(u)\). The derivative of \(\ln(u)\) with respect to \(u\) is \(\frac{1}{u}\). In our case: \(\frac{d}{d u}(\ln(u)) = \frac{1}{u}\).
03
Find the derivative of the inner function
Now, we need to find the derivative of the inner function, which is \(\ln(x)\). The derivative of \(\ln(x)\) with respect to \(x\) is \(\frac{1}{x}\). In our case: \(\frac{d}{d x}(\ln(x)) = \frac{1}{x}\).
04
Apply the chain rule
According to the chain rule, the derivative of the composite function is the product of the derivatives of the outer and inner functions. In our case, we have:
\(\frac{d}{d x}(\ln(\ln(x))) = \frac{d}{d u}(\ln(u))\cdot\frac{d}{d x}(u)\)
05
Substitute the values of the derivatives
Now, we can substitute the values of the derivatives of the outer function and inner function into the equation:
\(\frac{d}{d x}(\ln(\ln(x))) = \frac{1}{u}\cdot\frac{1}{x}\)
06
Replace u with the original expression
We set \(u = \ln(x)\) in step 1. Now replace \(u\) with \(\ln(x)\) back in the previous equation:
\(\frac{d}{d x}(\ln(\ln(x))) = \frac{1}{\ln(x)}\cdot\frac{1}{x}\)
07
Simplify the expression
The final expression for the derivative of the given function is:
\(\frac{d}{d x}(\ln(\ln(x))) = \frac{1}{x \ln(x)}\)
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Calculus
Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. This field constitutes a major part of modern math education. It has two major branches, differential calculus (concerning rates of change and slopes of curves) and integral calculus (concerning accumulation of quantities and the areas under and between curves).
Differential calculus introduces the concept of the derivative, which measures how a function changes as its input changes. Derivatives are fundamental for solving problems in physics, engineering, economics, and beyond, giving an exact way to quantify how one variable depends on another.
Differential calculus introduces the concept of the derivative, which measures how a function changes as its input changes. Derivatives are fundamental for solving problems in physics, engineering, economics, and beyond, giving an exact way to quantify how one variable depends on another.
Natural Logarithm Derivatives
The derivative of the natural logarithm function (ln) has profound importance in calculus. The natural logarithm of a variable x is denoted as \(\ln(x)\), and its derivative with respect to x is given by \(\frac{d}{dx}(\ln(x)) = \frac{1}{x}\).
This formula is derived from the definition of \(e\), the base of the natural logarithms, as the unique number whose natural logarithm is 1. Therefore, the slope of the tangent line to the curve \(y = \ln(x)\) at any point is the reciprocal of the x-coordinate at that point.
This formula is derived from the definition of \(e\), the base of the natural logarithms, as the unique number whose natural logarithm is 1. Therefore, the slope of the tangent line to the curve \(y = \ln(x)\) at any point is the reciprocal of the x-coordinate at that point.
Chain Rule
The chain rule is a fundamental theorem in calculus used for finding the derivative of composite functions. When we have two functions composed together, such as \(f(g(x))\), the chain rule allows us to differentiate this function by taking the derivative of the outer function (f) with respect to the inner function (g), and then multiplying it by the derivative of the inner function (g) with respect to x.
In mathematical terms, if \(h(x) = f(g(x))\), then \(h'(x) = f'(g(x)) \cdot g'(x)\). The chain rule enables complex derivatives to be taken quickly and accurately, which is especially useful in scenarios where functions are nested within each other.
In mathematical terms, if \(h(x) = f(g(x))\), then \(h'(x) = f'(g(x)) \cdot g'(x)\). The chain rule enables complex derivatives to be taken quickly and accurately, which is especially useful in scenarios where functions are nested within each other.
Composite Functions
Composite functions occur when one function is applied within another function, effectively 'nesting' one function inside the other. This is denoted as \(f(g(x))\) where \(g\) is a function whose range is the domain of another function \(f\).
Understanding composite functions is crucial when applying the chain rule for derivatives. It is often helpful to think of composite functions as a process where the output of the inner function \(g(x)\) becomes the input for the outer function \(f\). In our exercise example, \(\ln(\ln(x))\) is a composite function with \(\ln(x)\) nested inside another logarithmic function.
Understanding composite functions is crucial when applying the chain rule for derivatives. It is often helpful to think of composite functions as a process where the output of the inner function \(g(x)\) becomes the input for the outer function \(f\). In our exercise example, \(\ln(\ln(x))\) is a composite function with \(\ln(x)\) nested inside another logarithmic function.
