Question

Compute the derivative of the given function. $$f(x)=\sin \left(\sin ^{-1} x\right)$$

Short answer

The derivative of the function is 1 for \(-1 \leq x \leq 1\).

Step-by-step solution

Understand the Function

The function is \(f(x) = \sin( \sin^{-1}(x) )\). Recognize that \(\sin^{-1}(x)\) is the inverse sine function, also known as arcsine.

Apply the Inverse Property

Since the sine function \(\sin\) and the inverse sine function \(\sin^{-1}\) are inverses of each other, \(\sin(\sin^{-1}(x)) = x\) for \(-1 \leq x \leq 1\). Thus, \(f(x) = x\) within this interval.

Find the Derivative

We know \(f(x) = x\), meaning the derivative \(f'(x)\) is the derivative of \(x\), which is 1. Thus, \(f'(x) = 1\) for \(-1 \leq x \leq 1\).

Key concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are essential tools in calculus, allowing us to work backwards from trigonometric values to angles. These functions include inverse sine (\(\sin^{-1}(x)\)), inverse cosine (\(\cos^{-1}(x)\)), and inverse tangent (\(\tan^{-1}(x)\)), among others. Each of these functions provides the corresponding angle whose trigonometric value is the input.

For example:

• The function \(\sin^{-1}(x)\), gives an angle \(\theta\) such that \(\sin(\theta) = x\).
• These functions typically have domains and ranges that are bounded, ensuring they remain true functions (each input corresponds to one output). For instance, the domain of \(\sin^{-1}(x)\) is \([-1, 1]\) and the range is \([-\frac{\pi}{2}, \frac{\pi}{2}]\).

When dealing with inverse trigonometric functions in calculus, it's important to remember their derivative forms.
This allows the differentiation of more complex expressions that include these inverse functions, such as when they're nested within other functions like sine, cosine, or tangent. Be sure to remember these functions when solving calculus problems!

Arcsine Function

The arcsine function, denoted as \(\sin^{-1}(x)\) or arcsin(x), is a specific inverse trigonometric function. It is used to find the angle whose sine is a given number. For instance, if \(x = \sin(\theta)\), then \(\theta = \sin^{-1}(x)\).

Key characteristics of the arcsine function include:

• The domain is \([-1, 1]\), meaning it only applies to inputs within this range.
• The range extends from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), covering angles in the first and fourth quadrants.

When \(\sin^{-1}(x)\) is applied to \(\sin(x)\), like \(\sin(\sin^{-1}(x))\), it results in simplifying the expression directly to \(x\), since they are inverse operations of each other.
This property significantly simplifies problems in calculus by reducing them to basic algebra, as seen when finding derivatives. Understanding this function's behavior is essential, especially in solving equations involving trigonometric identities.

Differentiation Rules

Differentiation is a fundamental calculus concept, used to find the rate at which a function changes. Several rules exist to assist in differentiating various types of functions efficiently, including product, quotient, and chain rules.

When dealing with simple functions, often the basic rules are applied directly, such as:

• The derivative of \(x\) is \(1\), indicating a constant rate of change.
• The power rule states that the derivative of \(x^n\) is \(nx^{n-1}\).

In the context of inverse trigonometric functions, such as arcsine, specialized derivatives come into play. For example:

• The derivative of \(\sin^{-1}(x)\) is: \[ \frac{1}{\sqrt{1-x^2}} \] when \(-1 < x < 1\).

By understanding these differentiation rules, complex calculus problems can be approached with confidence. Moreover, recognizing how inverse functions interact, such as \(\sin(\sin^{-1}(x))\), allows for simplifying methods that reduce potential errors in problem-solving.