SOLVED:Simplify The Following Expressions: (a) \sinh \ln X (b) \cosh ...

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Simplify the following expressions: (a) $\sinh \ln x$ (b) $\cosh \ln x, \quad$ (c) $\tanh \ln x$ (d) $\frac{\cosh \ln x+\sinh \ln x}{\cosh \ln x-\sinh \ln x}$ Simplify the following expressions: (a) $\sinh \ln x$(b) $\cosh \ln x, \quad$ (c) $\tanh \ln x$(d) $\frac{\cosh \ln x+\sinh \ln x}{\cosh \ln x-\sinh \ln x}$ Calculus: A Complete Course Robert Adams ,… 9th Edition Chapter 3, Problem 7 View All Chapters

Step 1

Substituting $\ln x$ for $x$, we get $\sinh(\ln x) = \frac{e^{\ln x} - e^{-\ln x}}{2}$. Show more…

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Submit Thanks for your feedback! Simplify the following expressions: (a) $\sinh \ln x$ (b) $\cosh \ln x, \quad$ (c) $\tanh \ln x$ (d) $\frac{\cosh \ln x+\sinh \ln x}{\cosh \ln x-\sinh \ln x}$ Play audio Feedback Powered by NumerAI

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Key Concepts

Hyperbolic Functions Definition This concept involves expressing hyperbolic functions, such as sinh and cosh, in terms of exponential functions. Specifically, sinh u = (e^u ? e^(?u))?2 and cosh u = (e^u + e^(?u))?2. These definitions enable the transformation of hyperbolic expressions into forms that can be more easily manipulated and simplified. Exponential and Logarithm Relationships Understanding the relationship between exponential and logarithm functions is crucial. Since the logarithm is the inverse of the exponential function, properties such as e^(ln x) = x and e^(?ln x) = 1/x are used to directly replace and simplify expressions where the argument of the hyperbolic function is a logarithm. Simplification Techniques in Algebra This refers to the methods used for algebraic manipulation and simplification of expressions. By substituting definitions and applying exponential properties, one can reduce complex expressions involving hyperbolic functions and logarithms to more basic forms, often revealing cancellations or well-known identities.

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Transcript

00:01 Okay, in this problem, we want to simplify these expressions by using the definitions. 00:06 Hybabolic sign is e to the stuff minus e to the minus stuff all over two. 00:17 Okay, so that's e to the lnx minus minus. 00:20 I'm going to write this as e to the lnx to the minus one all over two. 00:27 So that's x minus 1 over x all over 2. 00:34 And so who's make those all into fractions and then i'm going to multiply everything by the common denominator which is x that gives me x squared minus 1 over 2x now if you do the same thing with the hyperbolic cosine the only difference is there will be a plus sign in there so let's just write that one that'll be x squared plus 1 over 2x then the hyperbolic tangent of the ln is the hyperbolic sign over the hyperbolic cosine. 01:17 So that'll be x squared minus 1 over 2x over x squared plus 1 over 2x. 01:24 Multiply the top and bottom by 2x. 01:26 Those are gone... Need help? Use Ace Ace is your personal tutor. It breaks down any question with clear steps so you can learn. Start Using Ace Ace is your personal tutor for learning Step-by-step explanations Instant summaries Summarize YouTube videos Understand textbook images or PDFs Study tools like quizzes and flashcards Listen to your notes as a podcast

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