Problem 7 - Calculus Early Transcendentals

Question

Evaluate the derivatives of the following functions. $$f(x)=\sin ^{-1} 2 x$$

Step-by-Step Solution

Answer

Answer: The derivative of the function \(f(x)=\sin^{-1} 2x\) is \(f'(x) = \frac{2}{\sqrt{1-4x^2}}\).

Step 1: Identify the Functions#g_tag_content#The given function is a composite function, with the main function being the inverse sine function and the inner function being \(2x\). So, we have: - \(f(x) = \sin^{-1}x\) - \(g(x) = 2x\)

Step 2: Calculate the Derivative of Each Function#g_tag_content#Now, we need to find the derivatives of the functions \(f(x)\) and \(g(x)\): - Derivative of \(f(x)\): For inverse sine function, we know the derivative is $\frac{1}{\sqrt{1-x^']]) So, \(f'(x) = \frac{1}{\sqrt{1-x^2}}\) - Derivative of \(g(x)\): Since \(g(x) = 2x\), its derivative will be \(g'(x) = 2\)

Step 3: Apply the Chain Rule#g_tag_content#Now, we apply the chain rule to find the derivative of the overall function: $$\frac{d}{dx}(\sin^{-1}(2x)) = f'(g(x))\cdot g'(x)$$ Substitute the values we found in Step 2: $$\frac{d}{dx}(\sin^{-1}(2x)) = \frac{1}{\sqrt{1-(2x)^2}}\cdot 2$$

Step 4: Simplify the Expression#g_tag_content#Now, we simplify the expression: $$\frac{d}{dx}(\sin^{-1}(2x)) = \frac{2}{\sqrt{1-4x^2}}$$ So the derivative of the function \(f(x)=\sin^{-1} 2x\) is: $$f'(x) = \frac{2}{\sqrt{1-4x^2}}$$

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are the inverse processes of the trigonometric functions like sine, cosine, and tangent. These functions are crucial when we want to determine the angle that corresponds to a particular trigonometric value. In this exercise, we focus on the inverse sine function, which is denoted as \( \sin^{-1}(x) \) or \( \arcsin(x) \).
This function helps us find an angle whose sine is \( x \). Its derivative is an essential tool in calculus and is given by \( \frac{1}{\sqrt{1-x^2}} \). This makes it a unique function because it handles angles and requires careful attention to the input values.
The domain of \( \sin^{-1}(x) \) is restricted to make the function invertible, typically from \([-1, 1]\). Knowing the derivative of inverse trigonometric functions allows us to solve more complex problems involving rates of change of angles.

Chain Rule

The chain rule is a fundamental theorem in calculus used for differentiating composite functions. It states that the derivative of a composite function \( f(g(x)) \) is found by computing the derivative of the outer function \( f \) at the inner function \( g(x) \), and multiplying it by the derivative of the inner function \( g(x) \).
Mathematically, it is represented as:

• \( \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \)

The chain rule is particularly useful when dealing with nested functions.
In our exercise, applying the chain rule allows us to manage the composite nature of \( \sin^{-1}(2x) \). Here, the outer function is the inverse sine, and the inner function is \( 2x \). By utilizing their derivatives, we can find the derivative of the overall function smoothly.

Composite Functions

Composite functions involve the combination of two or more functions, where the output of one function becomes the input of another function. This can be represented as \( (f \circ g)(x) = f(g(x)) \).
Understanding composite functions is vital as they appear frequently in calculus and real-world applications.
In this exercise, \( \sin^{-1}(2x) \) is a composite function where the main function is \( \sin^{-1}(x) \) and the inner function is \( 2x \).
Recognizing the nature of composite functions allows us to use the chain rule effectively to differentiate them. Once you identify and separate the layers of functions, calculating derivatives becomes more manageable.