Problem 279 - Calculus Volume 1

Question

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 function given is an inverse trigonometric function: \( y = \sin^{-1}(x^2) \). This means we are dealing with the arcsine function of \( x^2 \).

Step 2: Differentiate using the chain rule

To find \( \frac{dy}{dx} \), we need to differentiate \( y \) with respect to \( x \). The derivative of \( \sin^{-1}(u) \) with respect to \( u \) is \( \frac{1}{\sqrt{1-u^2}} \). Let \( u = x^2 \), so the outer function is \( \sin^{-1}(u) \) and the inner function is \( u = x^2 \). We apply the chain rule: \[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \].

Step 3: Find \( \frac{du}{dx} \)

Differentiate the inner function \( u = x^2 \) with respect to \( x \). The derivative is \( \frac{du}{dx} = 2x \).

Step 4: Find \( \frac{dy}{du} \)

Differentiate the outer function with respect to \( u \): \( \frac{dy}{du} = \frac{1}{\sqrt{1-u^2}} \). Substitute \( u = x^2 \) into the equation, which becomes \( \frac{1}{\sqrt{1-(x^2)^2}} = \frac{1}{\sqrt{1-x^4}} \).

Step 5: Combine using chain rule

Using the chain rule: \[ \frac{dy}{dx} = \frac{1}{\sqrt{1-x^4}} \times 2x = \frac{2x}{\sqrt{1-x^4}} \].

Key Concepts

Understanding the Chain Rule

The chain rule is a fundamental tool in calculus for differentiating composite functions. When you have a function nested inside another function, such as in the expression \( y = \sin^{-1}(x^2) \), the chain rule allows us to find the derivative in steps.

• First, identify the outer function and the inner function. Here, the outer function is \( \sin^{-1}(u) \), and the inner function is \( u = x^2 \).
• The chain rule essentially states that the derivative of the composite function is the product of the derivative of the outer function with respect to the inner function, times the derivative of the inner function with respect to \( x \).

So using the chain rule, if \( \frac{dy}{du} \) is the derivative of the outer function and \( \frac{du}{dx} \) is the derivative of the inner function, then the overall derivative \( \frac{dy}{dx} \) can be found using:
\[\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}.\]This method breaks down a complex problem into manageable parts.

Exploring the Arcsine Function

The arcsine function, denoted as \( \sin^{-1}(x) \) or \( \arcsin(x) \), is the inverse of the sine function over a specific interval. The function returns the angle whose sine is a given number.

• It's important to remember that the range of \( \arcsin \) is limited to \( -\frac{\pi}{2} \leq y \leq \frac{\pi}{2} \).
• When differentiating the arcsine function, one should note that its derivative is given by \( \frac{1}{\sqrt{1-x^2}} \).

For the problem \( y = \sin^{-1}(x^2) \), when finding the derivative, the understanding is applied that you need to substitute the inner function \( u = x^2 \) into the derivative formula. Thus, the derivative calculation eventually becomes more than simply applying the derivative of the inverse function, due to composition with another function (thanks to the chain rule).

Step-by-Step Differentiation Process

Differentiating composite functions, especially those involving inverse trigonometric functions, requires a clear, step-by-step approach using the chain rule. Here's how the differentiation for \( y = \sin^{-1}(x^2) \) was accomplished.

• Step 1: Recognize the function. Here, you have \( y = \sin^{-1}(x^2) \).
• Step 2: Apply the chain rule. Calculate the derivative of the outer function \( \frac{dy}{du} = \frac{1}{\sqrt{1-u^2}} \) while knowing \( u = x^2 \).
• Step 3: Differentiate the inner function: \( u = x^2 \). This yields \( \frac{du}{dx} = 2x \).
• Step 4: Substitute \( u = x^2 \) into your \( \frac{dy}{du} \) derivative, resulting in \( \frac{1}{\sqrt{1-x^4}} \).
• Step 5: Combine these results using the chain rule: \( \frac{dy}{dx} = \frac{1}{\sqrt{1-x^4}} \times 2x = \frac{2x}{\sqrt{1-x^4}} \).

Mastering these steps helps students effectively differentiate similar functions, providing a solid understanding of calculus concepts.