Problem 22 - Calculus for Biology and Medicine

Question

Use (4.14) to find the derivative of the inverse at the indicated point. Denote the inverse of \(y=\sin x,-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}\), by \(y=\arcsin x\), \(-1 \leq x \leq 1\). Show that $$ \frac{d}{d x} \arcsin x=\frac{1}{\sqrt{1-x^{2}}}, \quad-1

Step-by-Step Solution

Answer

The derivative of \( \arcsin x \) is \( \frac{1}{\sqrt{1-x^2}} \) for \(-1

Step 1: Understanding the Inverse Function

The function given is \( y = \arcsin x \), which is the inverse of \( y = \sin x \) over the specified domain \( -\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \). To find the derivative of the inverse function \( y = \arcsin x \), we begin by recalling that for a differentiable inverse function, the derivative can be expressed using the formula for the derivative of an inverse function: \( (f^{-1})'(y) = \frac{1}{f'(x)} \) where \( x = f^{-1}(y) \).

Step 2: Express Derivative of Sine Function

For the function \( y = \sin x \), its derivative is \( \frac{d}{d x} (\sin x) = \cos x \). According to the inverse function theorem, if \( y = \arcsin x \), then \( x = \sin y \). The derivative of the arcsine function at a point \( x \) is \( \frac{1}{\cos y} \), where \( y = \arcsin x \).

Step 3: Relate \( \cos y \) to \( x \)

Use the Pythagorean identity \( \sin^2 y + \cos^2 y = 1 \) to express \( \cos y \) in terms of \( x \). Since \( \sin y = x \), we have \( \cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2} \). This is true in the domain \( -1 < x < 1 \) because \( \cos y \) is non-negative with \( -\frac{\pi}{2} \leq y \leq \frac{\pi}{2} \).

Step 4: Derive the Formula for \( \frac{d}{d x} \arcsin x \)

Substituting back into our formula for the derivative of the inverse function, we find \( \frac{d}{d x} \arcsin x = \frac{1}{\cos(\arcsin x)} = \frac{1}{\sqrt{1 - x^2}} \) for \( -1 < x < 1 \). This provides us the desired derivative expression for the arcsine function.

Key Concepts

Arcsin Derivative

The derivative of the arcsine function, known as the arcsin derivative, allows us to understand the rate at which the angle changes with respect to the input value. In the context of inverse trigonometric functions, specifically for the arcsin, we focus on determining this derivative. The standard expression for the derivative of arcsin is presented as:

• \( \frac{d}{dx} \arcsin(x) = \frac{1}{\sqrt{1 - x^2}} \)
• This formula is applicable within the domain \(-1 < x < 1\), ensuring the "output angle" is within the range \(-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}\).

To find this result, we employ the inverse function theorem, adapt the derivative of the sine function, and utilize the Pythagorean identity to reformulate expressions such as \(\cos(y)\) into terms of \(x\). This derivative demonstrates how gently or steeply the arcsine curve rises or falls, by indicating the slope at any given point on its graph.

Inverse Function Theorem

The inverse function theorem provides a systematic way to find derivatives of inverse functions in calculus. Specifically, when dealing with the inverse of a function such as \(y = \sin x\), understanding the theorem is essential. The theorem states:

• If \(y = f(x)\) is a differentiable function and its inverse \(x = f^{-1}(y)\) exists, the derivative of the inverse is given by:\[(f^{-1})'(y) = \frac{1}{f'(x)}\]

This crucial relationship is applied to determine the derivative of \(y = \arcsin x\). By setting \(x = \sin y\), we apply the inverse function theorem to find that the derivative of \(y = \arcsin x\) depends inversely on the derivative of \(y = \sin x\), which is \(\cos x\). Bringing in further understanding of trigonometric relationships aids in expressing \(\cos(y)\) in terms of \(x\). This application highlights how inverse relationships between functions lead us to novel derivative expressions.

Pythagorean Identity

The Pythagorean identity is a cornerstone of trigonometry, crucial for solving equations involving trigonometric functions. It states that:

• \( \sin^2 y + \cos^2 y = 1 \)

This identity not only helps in simplification but is vital when expressing one trigonometric function in terms of another. In the context of the arcsin derivative, we utilize the identity to express \(\cos y\) solely in terms of \(x\). Since \(\sin y = x\), by substituting, we get:

• \( \cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2} \)

This relationship is crucial above all because it allows substitution into the inverse derivative formula. By converting functions like \(\cos(y)\) into familiar forms with respect to \(x\), we simplify the arcsin expression into an easily computable derivative.

Trigonometric Functions

Trigonometric functions, including sine, cosine, and the inverse arcsine, are fundamental in describing periodic phenomena. These functions have unique properties and domains that allow us to transform geometric relationships into algebraic equations. Let’s outline these functions:

• Sine: Represents opposite/hypotenuse, defined over \(-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}\).
• Cosine: Represents adjacent/hypotenuse, providing derivatives like those found in inverse functions.
• Arcsine: The inverse of sine, restricted to \(-1 \leq x \leq 1\), giving angles as output.

Each function has its unique behavior but remains interconnected through identities like the Pythagorean identity. Knowing how these functions interact allows seamless conversion between angles and ratios, which is key in calculus when finding derivatives. This deep understanding reflects how changes in one aspect of a triangle impact others within a cyclical pattern of trigonometric curves.

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Other exercises in this chapter

Problem 22

(a) Use the formal definition to find the derivative of \(y=\) \(x^{2}+1\) at \(x=1\) (b) Show that the point \((1,2)\) is on the graph of \(y=x^{2}+1\), and fi

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Problem 22

Find the derivative with respect to the independent variable. $$ f(x)=-5 \cos \left(2-x^{3}\right)+2 \cos ^{3}(x-4) $$

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Problem 22

Differentiate the functions with respect to the independent variable. \(f(x)=\cos \left(e^{x}\right)\)

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Problem 22

Apply the product rule to find the normal line, in slope-intercept form, of \(y=f(x)\) at the specified point. \(f(x)=(1-x)\left(2-x^{2}\right)\), at \(x=2\)

View solution

Key Concepts

Arcsin Derivative

The derivative of the arcsine function, known as the arcsin derivative, allows us to understand the rate at which the angle changes with respect to the input value. In the context of inverse trigonometric functions, specifically for the arcsin, we focus on determining this derivative. The standard expression for the derivative of arcsin is presented as:

• \( \frac{d}{dx} \arcsin(x) = \frac{1}{\sqrt{1 - x^2}} \)
• This formula is applicable within the domain \(-1 < x < 1\), ensuring the "output angle" is within the range \(-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}\).

To find this result, we employ the inverse function theorem, adapt the derivative of the sine function, and utilize the Pythagorean identity to reformulate expressions such as \(\cos(y)\) into terms of \(x\). This derivative demonstrates how gently or steeply the arcsine curve rises or falls, by indicating the slope at any given point on its graph.

Inverse Function Theorem

The inverse function theorem provides a systematic way to find derivatives of inverse functions in calculus. Specifically, when dealing with the inverse of a function such as \(y = \sin x\), understanding the theorem is essential. The theorem states:

• If \(y = f(x)\) is a differentiable function and its inverse \(x = f^{-1}(y)\) exists, the derivative of the inverse is given by:\[(f^{-1})'(y) = \frac{1}{f'(x)}\]

This crucial relationship is applied to determine the derivative of \(y = \arcsin x\). By setting \(x = \sin y\), we apply the inverse function theorem to find that the derivative of \(y = \arcsin x\) depends inversely on the derivative of \(y = \sin x\), which is \(\cos x\). Bringing in further understanding of trigonometric relationships aids in expressing \(\cos(y)\) in terms of \(x\). This application highlights how inverse relationships between functions lead us to novel derivative expressions.

Pythagorean Identity

The Pythagorean identity is a cornerstone of trigonometry, crucial for solving equations involving trigonometric functions. It states that:

• \( \sin^2 y + \cos^2 y = 1 \)

This identity not only helps in simplification but is vital when expressing one trigonometric function in terms of another. In the context of the arcsin derivative, we utilize the identity to express \(\cos y\) solely in terms of \(x\). Since \(\sin y = x\), by substituting, we get:

• \( \cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2} \)

This relationship is crucial above all because it allows substitution into the inverse derivative formula. By converting functions like \(\cos(y)\) into familiar forms with respect to \(x\), we simplify the arcsin expression into an easily computable derivative.

Trigonometric Functions

Trigonometric functions, including sine, cosine, and the inverse arcsine, are fundamental in describing periodic phenomena. These functions have unique properties and domains that allow us to transform geometric relationships into algebraic equations. Let’s outline these functions:

• Sine: Represents opposite/hypotenuse, defined over \(-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}\).
• Cosine: Represents adjacent/hypotenuse, providing derivatives like those found in inverse functions.
• Arcsine: The inverse of sine, restricted to \(-1 \leq x \leq 1\), giving angles as output.

Each function has its unique behavior but remains interconnected through identities like the Pythagorean identity. Knowing how these functions interact allows seamless conversion between angles and ratios, which is key in calculus when finding derivatives. This deep understanding reflects how changes in one aspect of a triangle impact others within a cyclical pattern of trigonometric curves.