Problem 25 - Thomas Calculus

Question

In Exercises \(25-36,\) find the derivative of \(y\) with respect to the appropriate variable. $$y=\sinh ^{-1} \sqrt{x}$$

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = \frac{1}{2\sqrt{x(x+1)}}\)

Step 1: Understand the Function

The given function is \(y = \sinh^{-1}(\sqrt{x})\), which involves the inverse hyperbolic sine function applied to \(\sqrt{x}\). Our goal is to find \(\frac{dy}{dx}\).

Step 2: Recall the Derivative Formula for Inverse Hyperbolic Sine

The derivative of \(\sinh^{-1}(u)\) with respect to \(u\) is \(\frac{1}{\sqrt{u^2 + 1}}\). We will use this formula, but recognize we have \(u = \sqrt{x}\).

Step 3: Apply the Chain Rule

To differentiate \(y = \sinh^{-1}(\sqrt{x})\), apply the chain rule. Differentiate the outside function \(\sinh^{-1}(u)\) with respect to \(u = \sqrt{x}\), and then multiply by the derivative of \(u = \sqrt{x}\) with respect to \(x\).

Step 4: Differentiate the Inside Function

The inside function is \(u = \sqrt{x} = x^{1/2}\). The derivative of \(u\) with respect to \(x\) is \(\frac{1}{2}x^{-1/2}\).

Step 5: Combine the Derivatives

Substitute \(u = \sqrt{x}\) into the derivative formula for \(\sinh^{-1}(u)\):\[ \frac{d}{dx} [\sinh^{-1}(\sqrt{x})] = \frac{1}{\sqrt{(\sqrt{x})^2 + 1}} \cdot \frac{1}{2}x^{-1/2} \]Simplify the expression: \((\sqrt{x})^2 = x\), so:\[ \frac{dy}{dx} = \frac{1}{\sqrt{x + 1}} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{2\sqrt{x(x+1)}} \]

Step 6: Simplify the Expression

The expression simplifies to:\[ \frac{dy}{dx} = \frac{1}{2\sqrt{x(x+1)}} \]This is the simplified form of the derivative of \(y = \sinh^{-1}(\sqrt{x})\) with respect to \(x\).

Key Concepts

Inverse Hyperbolic Functions

Inverse hyperbolic functions are similar to inverse trigonometric functions but are defined using hyperbolic functions like \(\sinh, \cosh, \,\text{and}\, \tanh\). They have significant applications in advanced mathematics, physics, and engineering. The inverse hyperbolic sine function, denoted as \(\sinh^{-1}(x)\), is particularly essential, as it allows us to solve equations involving hyperbolic sines. Unlike regular trigonometric functions, hyperbolic functions deal with hyperbolas rather than circles, conferring unique properties to their inverses.
A key property of inverse hyperbolic functions is how they relate to logarithms. Specifically, the inverse hyperbolic sine can be expressed as a logarithm: \(\sinh^{-1}(x) = \ln(x + \sqrt{x^2 + 1})\). This representation shows the function's real-valued domain and smoothness, making it easy to differentiate, as needed in calculus problems. When you encounter these functions in problems, it's helpful to remember their logarithmic counterparts for deeper understanding.

Chain Rule

The chain rule is a cornerstone of differential calculus, especially when dealing with composite functions, where one function is inside another. This rule provides a technique to compute the derivative of such functions efficiently.
In its basic form, the chain rule states: if you have \(y = f(g(x))\), then the derivative of \(y\) with respect to \(x\) is given by \((f'(g(x)) \cdot g'(x))\). This essentially requires you to take the derivative of the outer function, evaluated at the inner function, and multiply it by the derivative of the inner function itself.
Employing the chain rule in our initial problem with \(y = \sinh^{-1}(\sqrt{x})\) involves recognizing \(g(x) = \sqrt{x} \, \text{and}\, f(u) = \sinh^{-1}(u)\). By first taking the derivative of \(\sinh^{-1}(u)\) and then multiplying by the derivative of \(\sqrt{x}\), we achieve the derivative sought, showcasing the versatility of this rule.

Differential Calculus

Differential calculus revolves around the concept of a function's rate of change. It allows us to determine how a function changes as its inputs change, a principle captured through derivatives. Derivatives measure instantaneous rates of change, slopes of tangent lines, and are foundational in more complex analysis.
The main operation in differential calculus is differentiation, the process of finding a derivative. This involves several rules and methods, such as the product rule, quotient rule, and the chain rule, used to differentiate complicated functions effectively. Moreover, understanding the geometric interpretation of derivatives—how they reflect the steepness or flatness of a curve at certain points—is crucial for deeper mathematical comprehension.
In practical terms, differential calculus enables problem-solving in areas like physics for movement analysis, economics for optimizing functions, and beyond. By mastering differentiation techniques, you can approach various real-world problems with logical precision.