Problem 22 - Calculus Single Variable

Question

Differentiate the given expression with respect to \(x\). $$ \arccos \left(x^{2}\right) $$

Step-by-Step Solution

Answer

The derivative is \( \frac{-2x}{\sqrt{1-x^4}} \).

Step 1: Identify the function to differentiate

The given function is \( f(x) = \arccos(x^2) \). Our task is to find the derivative \( f'(x) \).

Step 2: Use chain rule for differentiation

To differentiate \( \arccos(u) \), where \( u = x^2 \), we use the chain rule: \( \frac{d}{dx}[\arccos(u)] = \frac{-1}{\sqrt{1-u^2}} \cdot \frac{du}{dx} \).

Step 3: Differentiate the inner function

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

Step 4: Apply the chain rule result

Substitute \( u = x^2 \) and \( \frac{du}{dx} = 2x \) into the chain rule: \[ \frac{d}{dx}[\arccos(x^2)] = \frac{-1}{\sqrt{1-(x^2)^2}} \cdot 2x = \frac{-2x}{\sqrt{1-x^4}}. \]

Step 5: Simplify the expression

The derivative of \( \arccos(x^2) \) with respect to \( x \) is \( \frac{-2x}{\sqrt{1-x^4}} \).

Key Concepts

Chain Rule in Calculus

The chain rule is a fundamental concept in calculus, primarily used when differentiating composite functions. Imagine peeling back the layers of an onion: that's what the chain rule helps you do when dealing with functions wrapped around others. The idea is simple. You differentiate the outer function, and then multiply it by the derivative of the inner function.

Here's the formula at its core: if you have a composite function \( f(g(x)) \), then the derivative \( f'(x) \) is: \ \[ f'(x) = f'(g(x)) \cdot g'(x). \]

• Start with differentiating the outside function.
• Next, differentiate the inside function.
• Multiply them together.

This rule is immensely handy when dealing with layers of complexity in functions, keeping things manageable and structured.

Inverse Trigonometric Functions

Inverse trigonometric functions, such as arc cosine (\( \arccos \)), allow us to find angles when given a ratio of sides in a right-angled triangle. For example, \( \arccos(x) \) gives the angle whose cosine is \( x \). But when it comes to calculus and differentiation, these functions have their behavior and properties. Especially in how they respond to differentiation.

For \( \arccos(u) \), the derivative is: \ \[ \frac{d}{dx}[\arccos(u)] = \frac{-1}{\sqrt{1-u^2}} \frac{du}{dx}. \]
This formula shows that the rate of change of an inverse trig function involves an adjustment by the inner function's derivative. The presence of that square root in the denominator is crucial, as it handles the constraint where the input values need to fall between -1 and 1.

• \( \arccos \) provides angles.
• When differentiating, you introduce a negative factor.
• The expression \( \sqrt{1-u^2} \) ensures valid inputs.

These properties and behaviors are essential when working through differentiation problems.

Understanding Derivatives of Composite Functions

Composite functions are built by stacking one function inside another, a process that often occurs in mathematics. To differentiate such functions, the chain rule becomes the go-to method, providing a structured process to peel back complexity.

Consider a function like \( \arccos(x^2) \). It combines the inverse trigonometric function \( \arccos \) and the quadratic function \( x^2 \). To find the derivative, you need to:

• Differentiate the outer layer, \( \arccos(u) \), using the inverse trig formula.
• Find the derivative of the inner function, \( u = x^2 \).
• Apply the chain rule with these derivatives.

This combination of techniques summarizes the approach to differentiating composite functions. Each piece plays a specific role, ensuring we respect the unique characteristics of each function involved.