Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. \(\frac{d}{d x}(\sinh (\ln 3))=\frac{\cosh (\ln 3)}{3}\) b. \(\frac{d}{d x}(\sinh x)=\cosh x\) and \(\frac{d}{d x}(\cosh x)=-\sinh x\) c. \(\ln (1+\sqrt{2})=-\ln (-1+\sqrt{2})\) d. \(\int_{0}^{1} \frac{d x}{4-x^{2}}=\frac{1}{2}\left(\operatorname{coth}^{-1} \frac{1}{2}-\operatorname{coth}^{-1} 0\right)\)

Short answer

a. The statement is false, as the derivative of sinh(ln(3)) with respect to x is 0. b. The first statement is true, as the derivative of sinh(x) is cosh(x). The second statement is false, as the derivative of cosh(x) is sinh(x), not -sinh(x). c. The statement is false, as the logarithm of a negative number is not valid for real logarithm functions, and -1 + sqrt(2) is a positive number. d. The statement is false, as the integration of the given function is equal to (1/2)(coth^(-1)(1/sqrt(2)) - coth^(-1)(0)), which is not equal to the given answer.

Step-by-step solution

a. Determine the truth of the statement about the derivative of sinh(ln(3))

Differentiate \({\sinh(\ln(3))}\) with respect to \(x\) using Chain Rule: \(\frac{d}{dx}(\sinh(u))\bigg|_{u=\ln(3)} \cdot \frac{du}{dx}\), where \(u = \ln(3)\). Since \(\frac{d}{dx}(\sinh(u))=\cosh(u)\), we have \(\frac{d}{dx}(\sinh(u))\bigg|_{u=\ln(3)}=\cosh(\ln(3))\). However, \(\frac{du}{dx}\), where \(u=\ln(3)\), is a constant, and its derivative is zero. Thus, \(\frac{d}{dx}(\sinh(\ln(3))) = 0\). The given statement is false.

b. Determine the truth of the statements about the derivatives of sinh(x) and cosh(x)

Differentiate \({\sinh(x)}\) with respect to \(x\) using the definition of sinh: \(\sinh(x)=\frac{e^x-e^{-x}}{2}\). Using the Chain Rule: \(\frac{1}{2}(e^x + e^{-x}) = \cosh(x)\). Thus the first statement is true. Next, differentiate \({\cosh(x)}\) with respect to \(x\) using the definition of cosh: \(\cosh(x)=\frac{e^x+e^{-x}}{2}\). Using the Chain Rule: \(\frac{1}{2}(e^x - e^{-x}) = \sinh(x)\). The second statement is false since the derivative of \(\cosh(x)\) is not \(-\sinh(x)\), but it is \(\sinh(x)\).

c. Determine the truth of the statement about logarithms

Recall that the logarithm of a negative number is not valid for real logarithm functions. Since \(-1+\sqrt{2}\) is positive, the expression \(\ln(-1+\sqrt{2})\) is not valid, and thus, the given statement is false.

d. Determine the truth of the statement about the integration and inverse hyperbolic functions

Integrate the function with respect to \(x\): \(\int_{0}^{1} \frac{dx}{4-x^2}=\int_{0}^{1} \frac{dy}{(2+y)(2-y)}\), where \(y=x^2/2\). Using partial fraction decomposition: \(\frac{1}{(2+y)(2-y)}=\frac{A}{2+y}+\frac{B}{2-y}\). Solve for A and B: \(A = \frac{1}{4}\) and \(B = -\frac{1}{4}\). Thus, the integral becomes: \(\int_{0}^{1} \left(\frac{1}{4(x^2+2)} -\frac{1}{4(2-x^2)}\right) dx\). Integrate each term separately: \(\frac{1}{2}(\int_{0}^{1}\frac{dx}{x^2+2} - \int_{0}^{1} \frac{dx}{2-x^2}) = \frac{1}{2}(\operatorname{coth}^{-1}(\frac{1}{\sqrt{2}}) - \operatorname{coth}^{-1}(0))\). Since the given statement is: \(\int_{0}^{1} \frac{dx}{4-x^2}=\frac{1}{2}(\operatorname{coth}^{-1}(\frac{1}{2})-\operatorname{coth}^{-1}(0))\), we can see that the statement is false, as the calculated answer is not equal to the given answer.

Key concepts

Derivative of Hyperbolic Functions

Hyperbolic functions such as sinh(x) and cosh(x) are analogs of the ordinary trigonometric functions, but for hyperbolas instead of circles. Understanding the derivatives of these functions is critical in solving calculus problems.

For instance, the derivative of \( \sinh(x) \) is \( \cosh(x) \) and not \( -\sinh(x) \), as you might initially think if you were drawing a parallel from the circular function \( \sin(x) \). This mistake is common among students who are new to calculus. It's essential to remember that while \( \sin(x) \) and \( \sinh(x) \) may look similar, they behave quite differently concerning their derivatives.

Chain Rule

The Chain Rule is a powerful differentiation tool that allows calculation of the derivative of composite functions. It's often used when a function is nested inside another function, like a Russian doll.

For example, if you have a function like \( \sinh(\ln(x)) \), you would first differentiate the outer function (sinh) with respect to its inner function (ln), and then multiply that result by the derivative of the inner function with respect to x. This approach extends to more complex functions, providing a systematic way to tackle differentiation. It's like unwrapping layers to find the core — each step requires careful unpacking to ensure accuracy.

Logarithm Properties

Logarithm properties can simplify complex expressions and make it easier to work with exponents. They are the 'handy tools' of algebra and pre-calculus.

Some essential properties to remember are the logarithm product rule, quotient rule, and power rule. For instance, the relationship \(\ln(a) + \ln(b) = \ln(a \times b)\) can turn multiplication within a logarithm into addition outside of it — a great trick for simplifying algebraic expressions. However, using these properties correctly requires careful attention to details such as the domain of the logarithmic functions to avoid invalid operations like taking the logarithm of a negative number.

Integral Calculus

Integral calculus deals with finding the area under curves and is the inverse process of differentiation. When working with integrals, especially definite integrals, you're essentially looking for the accumulated area from one point to another.

For definite integrals, such as \(\int_{a}^{b} f(x) \, dx\), you are finding the sum of infinitely small rectangles under the curve of f(x) from x = a to x = b. Integral calculus is like piecing together a puzzle — each infinitesimal piece contributes to the total picture. This concept is widely used in various fields, including physics, engineering, and even finance.

Partial Fraction Decomposition

Partial fraction decomposition is a technique to break down complex rational expressions into simpler ones that are easier to integrate or understand. It's akin to breaking a compound into simpler substances in chemistry.

The process involves expressing a fraction as a sum of fractions with simpler denominators. Solving the resulting system of equations will yield coefficients that, when put back together, will form the original complex expression in its decomposed form. This method is indispensable for integrating rational functions and is very much like solving a mystery, where you gradually unearth and piece together clues to arrive at the conclusion.

Inverse Hyperbolic Functions

Inverse hyperbolic functions, such as \(\operatorname{coth}^{-1}(x)\), undo what the original hyperbolic functions do. They are the 'reverse gear' of hyperbolic functions, taking you back to the angle whose hyperbolic function value you started with.

Understanding these inverse functions is key when dealing with integrals that result from partial fraction decomposition, or when you need to simplify complex expressions. Remember that inverse hyperbolic functions can also be expressed through logarithmic functions due to their similar properties. This linkage between different types of functions can often provide alternate avenues for solving difficult problems.