Problem 41 - Thomas Calculus in SI Units

Question

Find the derivative of \(y\) with respect to the appropriate variable. $$y=x \sin ^{-1} x+\sqrt{1-x^{2}}$$

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dx} = \sin^{-1}(x) \).

Step 1: Recognize the function

Identify that the function is given by \( y = x \sin^{-1}(x) + \sqrt{1 - x^2} \). This function combines a product and a composite of trigonometric inverse functions.

Step 2: Derive using the product rule

For the first term \( x \sin^{-1}(x) \), use the product rule. If \( u = x \) and \( v = \sin^{-1}(x) \), then \( \frac{dy}{dx} = u'v + uv' \). Calculate each derivative component.

Step 3: Differentiate \(x\)

Differentiate \( u = x \), which simply gives \( u' = 1 \).

Step 4: Differentiate \(\sin^{-1}(x)\)

For \( v = \sin^{-1}(x) \), the derivative is \( \frac{dv}{dx} = \frac{1}{\sqrt{1-x^2}} \).

Step 5: Apply the Product Rule

Now combine the derivatives from the product rule: \( \frac{dy}{dx} = (1)\sin^{-1}(x) + x\left(\frac{1}{\sqrt{1-x^2}}\right) \).

Step 6: Differentiate \(\sqrt{1-x^2}\)

Apply the chain rule to differentiate \( \sqrt{1-x^2} \), which involves rewriting as \( (1-x^2)^{1/2} \). The derivative is \( \frac{d}{dx} (1-x^2)^{1/2} = \frac{-x}{\sqrt{1-x^2}} \).

Step 7: Combine the derivatives

Combine the derivatives of both terms: \( \frac{dy}{dx} = \sin^{-1}(x) + \frac{x}{\sqrt{1-x^2}} - \frac{x}{\sqrt{1-x^2}} \).

Step 8: Simplify the expression

Notice the \( \frac{x}{\sqrt{1-x^2}} \) terms cancel out to simplify \( \frac{dy}{dx} = \sin^{-1}(x) \).

Key Concepts

Product Rule

The product rule is a fundamental concept in calculus used to find the derivative of a product of two functions. When you have a function of the form \( y = u(x)v(x) \), where both \(u\) and \(v\) are functions of \(x\), the derivative \( \frac{dy}{dx} \) is given by:

• \( u'(x)v(x) + u(x)v'(x) \)

This rule is essential because, unlike simple multiplication of numbers, multiplying functions involves the rates of change of both functions. To apply this rule successfully, differentiate each function separately:

• Find \( u'(x) \): the derivative of \( u(x) \) with respect to \( x \).
• Find \( v'(x) \): the derivative of \( v(x) \) with respect to \( x \).

Plug these derivatives back into the product rule formula.
In the given exercise, \( y = x \sin^{-1}(x) \), we identify \( u = x \) and \( v = \sin^{-1}(x) \). Differentiating gives us \( u' = 1 \) and \( v' = \frac{1}{\sqrt{1-x^2}} \). By plugging these into the product rule formula, you accurately capture the contribution of each part of the product to the derivative.

Trigonometric Inverse Functions

Trigonometric inverse functions like \( \sin^{-1}(x), \cos^{-1}(x), \tan^{-1}(x) \) are employed to find the angle when you know the ratio. The derivative of these inverse functions is different from regular trigonometric functions.
For \( \sin^{-1}(x) \), the derivative \( \frac{d}{dx}(\sin^{-1}(x)) \) is \( \frac{1}{\sqrt{1-x^2}} \). This derivative is crucial for understanding how the function changes relative to changes in \(x\), especially because it involves a square root, which suggests a restriction, \(-1 \leq x \leq 1\).
When working with these derivatives, ensure you understand the geometric meaning they convey. Each trigonometric inverse function translates a ratio into an angle, with derivatives that reflect how this angle changes with respect to \(x\). In practical applications, this allows you to model situations where angles change based on the inputs.

Chain Rule

The chain rule in calculus is a method for differentiating composite functions, i.e., functions within functions. It is articulated by the formula:

• \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \)

This rule helps in understanding how changing \(x\) affects \(y\) when \(y = f(g(x))\), a composition of two functions. The idea is to take the derivative of the outer function, treating the inner function as a constant, and multiply it by the derivative of the inner function.
In \( y = \sqrt{1-x^2} \), rewrite it as \((1-x^2)^{1/2}\) to see it more clearly as a composite function.

• The outer function is \( y = u^{1/2} \) where \( u = 1-x^2 \), so \( \frac{dy}{du} = \frac{1}{2}u^{-1/2} \).
• The inner function is \( u(x) = 1-x^2 \), so \( \frac{du}{dx} = -2x \).

Thus, the chain rule yields\[-\frac{x}{\sqrt{1-x^2}}\]. This approach underlines the power of the chain rule in dealing with layers of functions and also showcases its utility in calculating derivatives that appear complex at first glance.