Problem 267 - Calculus Volume 1

Question

Use the functions \(y=f(x)\) to find a. \(\frac{d f}{d x}\) at \(x=a\) and b. \(\quad x=f^{-1}(y)\) c. Then use part b. to find \(\frac{d f^{-1}}{d y}\) at \(y=f(a)\). $$ f(x)=\sin x, x=0 $$

Step-by-Step Solution

Answer

a. \( f'(0) = 1 \); b. \( x = 0 \); c. \( \frac{d f^{-1}}{d y} = 1 \).

Step 1: Differentiate the Function

First, we need to find the derivative of the function \( f(x) = \sin x \). The derivative is \( f'(x) = \cos x \). Since we need to evaluate the derivative at \( x = 0 \), we substitute \( x = 0 \) into the derivative to find \( f'(0) = \cos(0) = 1 \).

Step 2: Find the Inverse Function

The function given is \( f(x) = \sin x \). The inverse function, \( f^{-1}(y) \), is \( x = \arcsin(y) \). This means for any \( y \), \( x = \arcsin(y) \) is the value that makes \( \sin(x) = y \). When \( x = 0 \), \( y = f(0) = \sin(0) = 0 \). So, the inverse function related to \( y = 0 \) is \( x = \arcsin(0) = 0 \).

Step 3: Differentiate the Inverse Function

To find \( \frac{d f^{-1}}{d y} \), we need the derivative of the inverse function. Using the derivative of \( x = \arcsin(y) \), we know \( \frac{d}{dy} \arcsin(y) = \frac{1}{\sqrt{1-y^2}} \). We need \( \frac{d f^{-1}}{d y} \) at \( y = f(0) = 0 \). Therefore, \( \frac{d f^{-1}}{d y} \Bigr|_{y=0} = \frac{1}{\sqrt{1-0^2}} = 1 \).

Key Concepts

Differentiation

Differentiation is a key concept in calculus which involves finding the rate at which a function changes at any given point. It is represented by the derivative of a function. For a function \( f(x) \), its derivative is typically denoted as \( f'(x) \) or \( \frac{df}{dx} \). Differentiation helps us understand how a function behaves and is the foundation for many calculus applications.

To differentiate \( f(x) = \sin x \), we apply standard differentiation rules. The derivative of \( \sin x \) is \( \cos x \). This tells us the slope of the tangent line to the curve \( y = \sin x \) at any point \( x \).

When we want the value of the derivative at a specific point, like \( x = 0 \), we substitute this value into the derivative function. Thus, \( f'(0) = \cos(0) = 1 \). This means at \( x = 0 \), the slope of the tangent line is \( 1 \), indicating a linear increase.

Trigonometric Functions

Trigonometric functions are fundamental to mathematics and describe relationships between the angles and sides of triangles. The six basic trigonometric functions are sine, cosine, tangent, cotangent, secant, and cosecant. These functions find extensive use in various fields, including engineering, physics, and architecture.

The sine function, \( \sin x \), is periodic, with a period of \( 2\pi \), and varies between \(-1\) and \(1\). It represents the y-coordinate of a point on the unit circle, making it crucial in circular and oscillatory motion analysis.

In this exercise, we used the sine function and its properties to find its derivative. Recognizing these functions and understanding their properties allows us to effectively differentiate, integrate, and apply them to solve real-world problems.

Moreover, recognizing when to use inverse trigonometric functions such as \( \arcsin \) is important. It helps us solve equations where we know the output value and need to find the angle or input that produces it.

Derivative of Inverse Function

The derivative of an inverse function is useful in understanding how changes in the output of a function relate to changes in its input. If \( y = f(x) \), and \( x = f^{-1}(y) \), the relationship between their derivatives follows from the inverse function theorem.

According to this theorem, if \( f \) is differentiable and its derivative \( f'(x) \) is not zero at a point, then the inverse \( f^{-1} \) is differentiable at \( y = f(x) \), and its derivative is \( \frac{dx}{dy} = \frac{1}{f'(x)} \).

In our solved example, we have \( f(x) = \sin x \) and its derivative \( f'(x) = \cos x \), which at \( x = 0 \) equals \( 1 \). Therefore, \( \frac{d}{dx}(x = \arcsin(y)) \) at \( y = 0 \) becomes \( \frac{1}{1} = 1 \). This demonstrates that a change in \( y \) results in an equal, reciprocal change in \( x \), reinforcing the interconnectedness of function and inverse through their derivatives.