Chapter 6: Problem 11
Find \(d y / d x\). $$y=\operatorname{coth}(\ln x)$$
Short Answer
Expert verified
\( \frac{dy}{dx} = -\frac{\operatorname{csch}^2(\ln x)}{x} \).
Step by step solution
01
Understand the Problem
We need to find the derivative of the function \( y = \operatorname{coth}(\ln x) \). This involves using chain rule and differentiation of hyperbolic functions.
02
Differentiate the Outer Function
Recall that the derivative of \( \operatorname{coth}(u) \) is \( -\operatorname{csch}^2(u) \). Here, \( u = \ln x \). Thus, the outer derivative will be: \[-\operatorname{csch}^2(\ln x)\].
03
Differentiate the Inner Function
The inner function is \( u = \ln x \), whose derivative is \( \frac{d}{dx}(\ln x) = \frac{1}{x} \).
04
Apply the Chain Rule
By applying the chain rule, we multiply the derivative of the outer function by the derivative of the inner function. Thus, \( \frac{dy}{dx} = -\operatorname{csch}^2(\ln x) \cdot \frac{1}{x} \).
05
Combine the Results
Combine the results from steps 2 and 3, yielding the final expression for \( \frac{dy}{dx} \):\[-\frac{\operatorname{csch}^2(\ln x)}{x}\].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chain Rule
In calculus, the chain rule is a fundamental technique for finding the derivative of a composite function. Think of it this way: When a function is made up of several nested functions, the chain rule helps us differentiate it by "pulling it apart" piece by piece. Here's how the chain rule works:
In our original exercise, we applied the chain rule to differentiate \(y = \operatorname{coth}(\ln x)\). We treated \(\operatorname{coth}()\) as our outer function and \(\ln x\) as our inner function, applying the rule step-by-step for accurate results.
- You first take the derivative of the outermost function and leave the inner function as it is.
- Then, you multiply this derivative by the derivative of the inside function.
In our original exercise, we applied the chain rule to differentiate \(y = \operatorname{coth}(\ln x)\). We treated \(\operatorname{coth}()\) as our outer function and \(\ln x\) as our inner function, applying the rule step-by-step for accurate results.
Hyperbolic Functions
Hyperbolic functions are analogs to the well-known trigonometric functions but based on hyperbolas instead of circles. These functions, such as \(\sinh(), \cosh(),\) and \(\operatorname{coth}()\), have important roles in various mathematical contexts.
To understand hyperbolic functions, it's useful to recognize their definitions:
These derivatives play crucial roles in the calculus of hyperbolic functions, as seen in our solution process where we differentiated \(y = \operatorname{coth}(\ln x)\). Understanding these functions enriches your calculus toolkit.
To understand hyperbolic functions, it's useful to recognize their definitions:
- \(\sinh(x) = \frac{e^x - e^{-x}}{2}\)
- \(\cosh(x) = \frac{e^x + e^{-x}}{2}\)
- \(\operatorname{coth}(x) = \frac{\cosh(x)}{\sinh(x)}\)
These derivatives play crucial roles in the calculus of hyperbolic functions, as seen in our solution process where we differentiated \(y = \operatorname{coth}(\ln x)\). Understanding these functions enriches your calculus toolkit.
Derivative of Natural Logarithm
The natural logarithm \(\ln(x)\) is a very special function in mathematics, commonly showing up in calculus due to its unique properties. Its derivative is one of the simplest you’ll encounter: \(\frac{d}{dx}(\ln x) = \frac{1}{x}\), for \(x > 0\). This reflects how the natural log function grows compared to linear functions.
In practice, the derivative \(\frac{1}{x}\) is often used in conjunction with other differentiation rules, like the chain rule. For instance, when differentiating \(y = \operatorname{coth}(\ln x)\), we used the derivative of \(\ln x\) as part of our chain rule application.
In practice, the derivative \(\frac{1}{x}\) is often used in conjunction with other differentiation rules, like the chain rule. For instance, when differentiating \(y = \operatorname{coth}(\ln x)\), we used the derivative of \(\ln x\) as part of our chain rule application.
- Recognizing when to utilize \(\frac{1}{x}\) can simplify the differentiation of potentially complex functions.