Problem 19 - Essential Calculus Early Transcendentals

Question

Find the derivative of the function. Simplify where possible. \(y=\sin ^{-1}(2 x+1)\)

Step-by-Step Solution

Answer

\(\frac{2}{\sqrt{-4x^2-4x}}\)

Step 1: Identify the Function Type

The given function is an inverse trigonometric function, specifically the inverse sine or arcsine function. We are looking for the derivative of \(y = \sin^{-1}(2x + 1)\).

Step 2: Recall the Derivative Rule for Inverse Sine

The derivative of \(\sin^{-1}(u)\) with respect to \(x\) is \(\frac{d}{dx}[\sin^{-1}(u)] = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}\). In this case, \(u = 2x + 1\).

Step 3: Differentiate "u" with Respect to "x"

Calculate \(\frac{du}{dx}\) for \(u = 2x + 1\). The derivative \(\frac{d}{dx}[2x + 1] = 2\).

Step 4: Apply the Chain Rule

Substitute \(u = 2x + 1\) and \(\frac{du}{dx} = 2\) into the derivative formula: \(\frac{d}{dx}[\sin^{-1}(2x + 1)] = \frac{1}{\sqrt{1-(2x+1)^2}} \cdot 2\).

Step 5: Simplify the Expression

Simplify the expression \(\sqrt{1-(2x+1)^2}\). First, expand \((2x+1)^2\) to get \(4x^2 + 4x + 1\). Thus, \(1 - (4x^2 + 4x + 1) = -4x^2 - 4x\), which simplifies the entire derivative to \(\frac{2}{\sqrt{-(4x^2 + 4x)}}\).

Step 6: State the Final Derivative Form

Thus, the derivative of \(y = \sin^{-1}(2x + 1)\) is \(\frac{2}{\sqrt{-4x^2-4x}}\).

Key Concepts

Derivatives

To understand the concept of derivatives, think of them as a way to determine the rate at which something changes. In calculus, finding a derivative means we're identifying how a function's output changes as its input changes. For a given function, the derivative is essentially the slope of the function at any point along its curve.
In our specific case, the function is an inverse trigonometric function, the arcsine of an expression. We're tasked with finding the derivative of this function, which will tell us how the output of the arcsine function changes as the variable changes. In general, when you find a derivative, you're uncovering a deeper understanding of how functions behave and interact with their variables.

Chain Rule

The chain rule is a fundamental concept in calculus that allows us to take derivatives of composite functions - functions within functions. Think of it as a set of instructions for peeling away the layers of an onion, each layer pertaining to a different function. When you have a function inside another function, as in our example \[ y = \sin^{-1}(2x + 1) \]The chain rule helps find the outer function's derivative by multiplying it with the derivative of the inner function.
For instance, in order to apply the chain rule effectively:

• Identify the outer function: Here, it's the arcsine function \( \sin^{-1}(u) \).
• Recognize the inner function: This is \( u = 2x + 1 \).
• Find the derivative of the outer function, \( \frac{1}{\sqrt{1-u^2}} \), and multiply it by the derivative of the inner function, \( \frac{du}{dx} = 2 \).

This method allows us to find the derivative of complex expressions by unraveling them step by step.

Arcsine Function

The arcsine function, written as \( \sin^{-1}(u) \) or 'inverse sine', is a type of inverse trigonometric function. It essentially undoes what the sine function does, taking a ratio and giving back the angle whose sine is that ratio.
In terms of derivatives, inverse trigonometric functions have their own unique set of rules. The derivative of the arcsine function is given by:\[ \frac{d}{dx}[\sin^{-1}(u)] = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx} \]For the given problem, we apply this rule by first individually deriving \( u = 2x + 1 \) leading to \( \frac{du}{dx} = 2 \), and then using the chain rule to bring it all together, which helps us handle the composite nature of the function.

Simplification

After taking the derivative, your expression might look complex. Simplification is the process of making it more manageable and easier to understand or compute. In our example, part of simplification involves dealing with the square root part:\[ \sqrt{1 - (2x + 1)^2} \]First, expand \((2x+1)^2\) to get \(4x^2 + 4x + 1\), then subtract it from 1:\[ 1 - (4x^2 + 4x + 1) = -4x^2 - 4x \]Now, the derivative expression becomes \( \frac{2}{\sqrt{-4x^2 - 4x}} \).Simplification not only reduces steps during calculations, but also aids in understanding the behavior of derivatives. In practice, it often involves algebraic manipulations to make expressions cleaner and easier to interpret.