Problem 279 - Calculus Volume 1

Question

For the following exercises, find \(\frac{d y}{d x}\) for the given function. $$y=\sin ^{-1}\left(x^{2}\right)$$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{2x}{\sqrt{1-x^4}} \)

Step 1: Identify the Function Type

The given function is \( y = \sin^{-1}(x^2) \), which is the inverse sine (arcsin) function. Recognize that we need to apply the derivative formula for the inverse sine function.

Step 2: Recall the Derivative Formula

The derivative of \( y = \sin^{-1}(u) \) with respect to \( x \) is \( \frac{d}{dx}[\sin^{-1} u] = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx} \). Here, \( u = x^2 \), so you will need to use the chain rule.

Step 3: Differentiate the Inner Function

First, differentiate \( u = x^2 \) with respect to \( x \). The derivative is \( \frac{du}{dx} = 2x \).

Step 4: Apply the Chain Rule

Substitute \( u = x^2 \) and \( \frac{du}{dx} = 2x \) into the derivative formula from Step 2. This gives us: \[ \frac{dy}{dx} = \frac{1}{\sqrt{1-(x^2)^2}} \cdot 2x. \]

Step 5: Simplify the Expression

Simplify the expression: \[ \frac{dy}{dx} = \frac{2x}{\sqrt{1-x^4}}. \] This is the final derivative of the function.

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are used to find angles when given a specific trigonometric value. These functions include \( \sin^{-1}, \cos^{-1}, \) and \( \tan^{-1} \), among others. Inverse functions essentially reverse the trigonometric process, turning ratios back into angles. In our exercise, \( y = \sin^{-1}(x^2) \) is the inverse sine function, also known as arcsine.The arcsine function specifically returns the angle whose sine is the given number. It is important to note that such functions have specific ranges to ensure they are appropriately defined as inverse functions. Thus, \( \sin^{-1}(x) \) has a range from \(-\pi/2\) to \(\pi/2\). These details ensure correct interpretation and application of inverse trigonometric functions. By using the derivative formula for \( \sin^{-1}(x) \), we can solve many calculus problems involving rate of change.

Chain Rule

The chain rule is a fundamental tool in calculus used for differentiating composite functions. A composite function is when one function is nested inside another, like \( y = \sin^{-1}(x^2) \). The chain rule helps us differentiate in these scenarios by breaking the process into manageable parts.The principle behind the chain rule is to differentiate the outer function, multiply by the derivative of the inner function. In our solution, we differentiated the arcsine function first and then multiplied it by the derivative of the inner function \( x^2 \) to get \( 2x \).Let's recall that if \( y = f(g(x)) \), the derivative \( \frac{dy}{dx} \) is found using: \[ \frac{dy}{dx} = f'(g(x)) \cdot g'(x). \] This makes the chain rule a powerful technique for dealing with complex derivatives, ensuring every component functions smoothly.

Differentiation Rules

Differentiation rules are specific methods used to find the derivative of functions. These rules simplify the process of computing derivatives, especially for more complex expressions. Some basic rules include the power rule, product rule, quotient rule, and chain rule. In this exercise, we highlight the derivative rule for inverse trigonometric functions.Each inverse trigonometric function has a derivative rule. For instance, the derivative of \( \sin^{-1}(u) \) is \( \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx} \). Knowing these rules allows you to handle a variety of functions with ease.By applying differentiation rules, mathematicians and students can swiftly work through various calculus problems. These rules are not only shortcuts but also essential tools for gaining deeper insights into the behavior of mathematical functions.