In Exercises 21–48, find the derivative of y with respect to the appropriate variable.
41. y= x arcsin(x) + √(1-x²)

Verified step by step guidance

1. Identify the function to differentiate: \(y = x \arcsin(x) + \sqrt{1 - x^2}\).
2. Recognize that the function is a sum of two parts, so use the sum rule: \(\frac{dy}{dx} = \frac{d}{dx}[x \arcsin(x)] + \frac{d}{dx}[\sqrt{1 - x^2}]\).
3. For the first part, \(x \arcsin(x)\), apply the product rule: \(\frac{d}{dx}[u v] = u' v + u v'\), where \(u = x\) and \(v = \arcsin(x)\).
4. Calculate the derivatives needed: \(u' = \frac{d}{dx}[x] = 1\) and \(v' = \frac{d}{dx}[\arcsin(x)] = \frac{1}{\sqrt{1 - x^2}}\).
5. For the second part, \(\sqrt{1 - x^2}\), rewrite it as \((1 - x^2)^{1/2}\) and use the chain rule: \(\frac{d}{dx}[(1 - x^2)^{1/2}] = \frac{1}{2}(1 - x^2)^{-1/2} \cdot (-2x)\).

Key Concepts

Derivative of Inverse Trigonometric Functions

The derivative of arcsin(x) is 1 / √(1 - x²). This rule is essential when differentiating expressions involving inverse sine functions, as it allows us to find the rate of change of arcsin(x) with respect to x.

Product Rule

The product rule is used to differentiate functions that are products of two differentiable functions. It states that (fg)' = f'g + fg', meaning the derivative of the product is the derivative of the first times the second plus the first times the derivative of the second.

Chain Rule

The chain rule is applied when differentiating composite functions. It states that the derivative of f(g(x)) is f'(g(x)) * g'(x). This is crucial for differentiating functions like √(1 - x²), where an inner function is nested inside an outer function.