Problem 23 - Thomas Calculus

Question

In Exercises \(21-42,\) find the derivative of \(y\) with respect to the appropriate variable. $$ y=\arcsin \sqrt{2} t $$

Step-by-Step Solution

Answer

The derivative of \( y \) with respect to \( t \) is \( \frac{\sqrt{2}}{\sqrt{1-2t^2}} \).

Step 1: Identify the Function Type

The function given is \( y = \arcsin(\sqrt{2} t) \). This is a composite function where the outer function is the \(\arcsin\) function and the inner function is \(\sqrt{2}t\). We need to apply the chain rule to find its derivative.

Step 2: Differentiate the Outer Function

To differentiate \( y = \arcsin(u) \), where \( u = \sqrt{2}t \), the derivative with respect to \( u \) is \( \frac{1}{\sqrt{1-u^2}} \). Apply this derivative rule to get \( \frac{1}{\sqrt{1-(\sqrt{2}t)^2}} \).

Step 3: Differentiate the Inner Function

The inner function \( u = \sqrt{2}t \) is linear in \( t \), so its derivative with respect to \( t \) is simply \( \sqrt{2} \).

Step 4: Apply the Chain Rule

According to the chain rule, the derivative of \( y \) with respect to \( t \) is given by multiplying the derivative of the outer function by the derivative of the inner function: \[ \frac{dy}{dt} = \frac{1}{\sqrt{1-(\sqrt{2}t)^2}} \times \sqrt{2} \].

Step 5: Simplify the Expression

Simplify the derivative expression: The derivative function becomes \[ \frac{dy}{dt} = \frac{\sqrt{2}}{\sqrt{1-2t^2}} \]. This is the derivative of \( y \) with respect to \( t \).

Key Concepts

Chain Rule

The chain rule is an essential tool in calculus for differentiating composite functions. It allows you to find the derivative of a function that is nested inside another function. The basic idea is to take the derivative of the outer function and multiply it by the derivative of the inner function. This is particularly useful when dealing with functions where one function is applied to the result of another function.

• When using the chain rule, identify the outer and inner functions first.
• Differentiate the outer function with respect to its inner function.
• Then, differentiate the inner function with respect to its variable.

Let's say you have a composite function \(f(g(x))\). The chain rule states that the derivative \(f'(g(x))\) must be multiplied by \(g'(x)\). Hence, the derivative is \(f'(g(x)) \cdot g'(x)\).
In our exercise, this rule helps us find \(\frac{dy}{dt}\) for \(y = \arcsin(\sqrt{2}t)\).

Composite Function

Composite functions are created when one function is situated inside another, forming a function composition. This means that the output of one function becomes the input of another.

• A composite function is written as \(f(g(x))\).
• The function \(g(x)\) is called the inner function.
• \(f(x)\) is known as the outer function.

In this exercise, \(y = \arcsin(\sqrt{2}t)\) is a composite function. The \(\arcsin()\) function acts as the outer function, while \(\sqrt{2}t\) is the inner function. Understanding how to work with composite functions is fundamental, as it helps apply the chain rule effectively.

Arcsin Function

The arcsin function is one of the inverse trigonometric functions, specifically the inverse of the sine function. It's denoted as \(\arcsin(x)\) or sometimes \(\sin^{-1}(x)\), and it returns the angle whose sine is \(x\).

• The domain of \(\arcsin(x)\) is \([-1,1]\), since those are the possible values for a sine function output.
• The range of \(\arcsin(x)\) is \([-\frac{\pi}{2},\frac{\pi}{2}]\), corresponding to the angles it can return.

When differentiating \(\arcsin(x)\), the derivative is \(\frac{1}{\sqrt{1-x^2}}\). It's crucial to apply this derivative correctly when dealing with composite functions like in our exercise. Here, you will handle \(\arcsin(\sqrt{2}t)\) by first considering the inner part \(\sqrt{2}t\) before substitution in its derivative.

Differentiation

Differentiation is the process of finding the derivative of a function, which represents how a function's output changes with varying input. It's a fundamental concept in calculus used to analyze and describe changes in functions.

• In context, differentiation gives us the slope or rate of change at a specific point on a curve.
• Derivatives are used extensively in various fields such as physics, economics, and engineering.

For the exercise, the function is \(y = \arcsin(\sqrt{2}t)\). Differentiating a composite function like this involves applying the chain rule as discussed earlier. You differentiate the outer function \(\arcsin(u)\) to get \(\frac{1}{\sqrt{1-u^2}}\), and then differentiate \(u = \sqrt{2}t\) with respect to \(t\), which gives \(\sqrt{2}\). Finally, multiply these results, ensuring that each step reflects the proper application of differentiation techniques.