Problem 21 - Calculus Single Variable

Question

Differentiate the given expression with respect to \(x\). $$ x \arcsin (x) $$

Step-by-Step Solution

Answer

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

Step 1: Apply the Product Rule

The expression given is a product of two functions: \( f(x) = x \) and \( g(x) = \arcsin(x) \). The product rule for differentiation states that \((uv)' = u'v + uv'\). Identify \(u = x\) and \(v = \arcsin(x)\).

Step 2: Differentiate Each Function

Differentiate \(u = x\) and \(v = \arcsin(x)\). The derivative of \(u = x\) is \(u' = 1\). The derivative of \(v = \arcsin(x)\) is \(v' = \frac{1}{\sqrt{1-x^2}}\).

Step 3: Apply the Product Rule

Substitute these derivatives into the product rule formula: \((u'v + uv') = (1) \cdot \arcsin(x) + x \cdot \frac{1}{\sqrt{1-x^2}}\).

Step 4: Simplify the Expression

Simplify the expression to obtain the final derivative: \( \arcsin(x) + \frac{x}{\sqrt{1-x^2}} \).

Key Concepts

Product Rule

When dealing with differentiation, the Product Rule is a powerful tool used when differentiating expressions that are the product of two separate functions. When we encounter a function composed of two multiplied functions, such as \( f(x) = u(x) \) and \( g(x) = v(x) \), the Product Rule becomes our ally. This rule dictates that the derivative of this product \((uv)\) can be calculated as:

• \((uv)' = u'v + uv'\)

Here, \(u'\) represents the derivative of \(u(x)\), and \(v'\) is the derivative of \(v(x)\).

Applying this rule involves three main steps:

• Differentiating \(u(x)\)
• Differentiating \(v(x)\)
• Combining these results using \((uv)' = u'v + uv'\)

This method ensures that each component of the product is properly considered in the differentiation process, allowing us to find the derivative accurately.

Inverse Trigonometric Functions

Inverse trigonometric functions are fascinating because they enable us to find angle measures based on trigonometric values. Essentially, they are the 'reverse' of the usual trigonometric functions like sine, cosine, and tangent. These functions are often denoted as \( \arcsin(x)\), \( \arccos(x)\), and \( \arctan(x)\).

For example, \( \arcsin(x)\) finds the angle whose sine value is \(x\). The range of \( \arcsin(x)\) is from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), meaning it outputs angles within this range to provide valid solutions.These functions are crucial in calculus, especially when dealing with problems involving integration and differentiation, as they help express relationships in a manner that's easier to manipulate mathematically. Moreover, their derivatives play a pivotal role when applying calculus operations to more complex functions.

Derivative of Arc Functions

The derivative of inverse trigonometric functions, or 'arc functions', is essential in calculus. When differentiating these functions, it's important to know how each one behaves and transforms.

Take \( \arcsin(x)\) as an example. The derivative of \( \arcsin(x)\) is:

• \( \frac{d}{dx}[\arcsin(x)] = \frac{1}{\sqrt{1-x^2}} \)

This derivative is defined so long as \(x\) is within the interval \([-1,1]\), ensuring the expression under the square root remains non-negative.Understanding these derivatives is critical when they appear as parts of larger expressions. When using them in calculations, such as applying the Product Rule, knowing these derivatives allows us to efficiently and accurately determine the behavior of composite functions.

Recognizing how these derivatives integrate with other calculus concepts can significantly simplify complex differentiation and integration problems.