Problem 25 - Thomas Calculus in SI Units

Question

Find the derivative of \(y\) with respect to the appropriate variable. $$y=\sinh ^{-1} \sqrt{x}$$

Step-by-Step Solution

Answer

The derivative of \( y \) with respect to \( x \) is \( \frac{1}{2\sqrt{x(1+x)}} \).

Step 1: Understanding the Function

The given function is \( y = \sinh^{-1} \sqrt{x} \), which means we need to find the derivative of the inverse hyperbolic sine function with respect to \( x \).

Step 2: Using the Derivative of Inverse Hyperbolic Sine

Recall that the derivative of \( \sinh^{-1}(u) \) with respect to \( u \) is \( \frac{1}{\sqrt{1 + u^2}} \). In our problem, \( u = \sqrt{x} \).

Step 3: Applying Chain Rule

Because \( u = \sqrt{x} = x^{1/2} \), use the chain rule for differentiation. First, find \( \frac{du}{dx} \), then apply it: \[ \frac{du}{dx} = \frac{1}{2\sqrt{x}} = \frac{1}{2}x^{-1/2} \].

Step 4: Differentiating \( y \) with Respect to \( x \)

Using the chain rule, the derivative \( \frac{dy}{dx} \) becomes: \[ \frac{dy}{dx} = \frac{1}{\sqrt{1 + u^2}} \cdot \frac{du}{dx} \]. Substitute \( u^2 = (\sqrt{x})^2 = x \): \[ \frac{dy}{dx} = \frac{1}{\sqrt{1 + x}} \cdot \frac{1}{2}x^{-1/2} \].

Step 5: Simplifying the Expression

Combine and simplify the expression: \[ \frac{dy}{dx} = \frac{1}{2\sqrt{x(1 + x)}} \]. This is the derivative of \( y \) with respect to \( x \).

Key Concepts

Chain Rule

When we find derivatives of composite functions, like our function involving inverse hyperbolic sine, the chain rule is an essential tool. The chain rule allows us to differentiate functions nested within others.
For a composite function like \( f(g(x)) \), the chain rule states: \[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \]

Let's apply this to our exercise where \( u = \sqrt{x} \). We're dealing with the inverse hyperbolic sine of \( u \), so we first need to get the derivative of \( \sinh^{-1}(u) \).
The process is:

• Differentiating \( \sinh^{-1}(u) \) gives \( \frac{1}{\sqrt{1 + u^2}} \).
• The chain rule tells us to multiply this by \( \frac{du}{dx} \).
• Here, \( \frac{du}{dx} \) is the derivative of \( \sqrt{x} \), calculated as \( \frac{1}{2\sqrt{x}} \).

This approach simplifies handling derivatives of inverse functions by breaking them into manageable parts.

Inverse Functions

Inverse functions reverse the operations performed by the original function. In our exercise, the function \( \sinh^{-1} \) is the inverse of the hyperbolic sine function \( \sinh \).

Understanding inverse functions is key in differentiation. When differentiating an inverse function, we often need the derivative of the original function. For hyperbolic functions:

• \( \sinh(x) \) has the inverse \( \sinh^{-1}(x) \).
• The derivative of \( \sinh^{-1}(x) \) is \( \frac{1}{\sqrt{1 + x^2}} \).

In our scenario, we're actually looking at \( \sinh^{-1}(\sqrt{x}) \). The idea is to recognize the simplification possible by knowing inverse relationships and how to navigate through their derivatives.

Differentiation Techniques

Differentiation techniques are crucial in calculus for finding the rate of change. With complex expressions like \( y = \sinh^{-1}(\sqrt{x}) \), multiple techniques often come into play.

Here's a closer look at our problem:

• Using basic rules: Start by recognizing that \( \sqrt{x} \) can be rewritten as \( x^{1/2} \), which simplifies differentiation.
• Chain rule: Apply the chain rule where \( u = \sqrt{x} \), differentiating \( u \), then using it to find \( \frac{dy}{dx} \).
• Simplification: Finally, simplify expressions, as seen when combining fractions: \( \frac{1}{\sqrt{1 + x}} \cdot \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x(1 + x)}} \).

These techniques together make handling derivatives of inverse hyperbolic functions possible when faced with complex expressions.