Problem 49 - Calculus of a Single Variable

Question

Find the derivative of the function. \(y=2 x \arccos x-2 \sqrt{1-x^{2}}\)

Step-by-Step Solution

Answer

The derivative of given function is \(2 * \arccos x - 4 * x / \sqrt{1 - x^{2}}\).

Step 1: Identify the type of function

Notice that there are two terms in the given function. The first term is a product of \(x\) and \(\arccos x\), thus applying the Product Rule, and the second term includes an element under the root, meaning the Chain Rule will be applied.

Step 2: Use the product rule for the first term

For the function \(f(x)g(x)\), the product rule is \(f'(x)g(x) + f(x)g'(x)\). Here, \(f(x) = x\) and \(g(x) = \arccos x\). So, \(f'(x) = 1\) and \(g'(x) = -1 / \sqrt{1 - x^{2}}\). Apply these to get \(1 *\arccos x + x*(-1 / \sqrt{1 - x^{2}})\). This simplifies to \(\arccos x - x / \sqrt{1 - x^{2}}\). Multiply by 2, the function becomes \(2 (\arccos x - x / \sqrt{1 - x^{2}})\).

Step 3: Use the chain rule for the second term

Note that the function underneath the square root, a is \(a = 1 - x^{2}\). Thus, applying the Chain Rule gives \(da/dx = -2x\), meaning the derivative of \(\sqrt{1 - x^{2}}\) is \(-(1/2) \sqrt{1 - x^{2}} (-2x)\), which simplifies to \(x / \sqrt{1 - x^{2}}\). Multiply this term by -2, yields \(-2 (x / \sqrt{1 - x^{2}})\).

Step 4: Combine and simplify

Combine steps 2 & 3. The function simplifies to \(2*\arccos x - 2 * x / \sqrt{1 - x^{2}} - 2*x / \sqrt{1 - x^{2}}\). Simplify to get \(2 * \arccos x - 4 * x / \sqrt{1 - x^{2}}\) as the derivative.

Key Concepts

Product Rule

The product rule is a fundamental principle used for differentiating products of two functions. If you have a function expressed as the product of two sub-functions, say \( f(x) \) and \( g(x) \), the product rule will help you find its derivative. The rule is expressed as: \[(fg)' = f'g + fg'\]This formula states that the derivative of the product is the derivative of the first function multiplied by the second function, plus the first function multiplied by the derivative of the second.

• Identify functions \( f(x) \) and \( g(x) \).
• Calculate \( f'(x) \) and \( g'(x) \).
• Apply the product rule formula.
• Simplify the expression if necessary.

Always remember to multiply through any constants outside the product to complete the differentiation.

Chain Rule

The chain rule is indispensable when dealing with composite functions, meaning functions nested inside one another. It allows you to find the derivative of a composed function by breaking it down into its parts. If you have a composite function \( h(x) = f(g(x)) \), you apply the chain rule as follows:\[ h'(x) = f'(g(x)) \cdot g'(x)\]This implies that you differentiate the outer function \( f \) with respect to the inner function \( g \), then multiply by the derivative of the inner function \( g \) itself.

• Express the function as a composition of two or more functions.
• Differentiate the outermost function while keeping the inside function unchanged.
• Multiply the result by the derivative of the inner function.

The chain rule is crucial when derivative terms contain square roots, trigonometric, or logarithmic functions where nesting is present.

Trigonometric Functions

Trigonometric functions, such as \( \sin, \cos, \tan \), have specific rules for derivatives that are essential for calculus. Understanding these allows you to work with periodic functions that model waves, circles, and oscillations. Here are the basic derivatives for some common trigonometric functions:

• \( \frac{d}{dx} \sin x = \cos x \)
• \( \frac{d}{dx} \cos x = -\sin x \)
• \( \frac{d}{dx} \tan x = \sec^2 x \)

When these functions are part of a more complex expression, you might need to use other rules like the product or chain rule to differentiate effectively.

Inverse Trigonometric Functions

Inverse trigonometric functions, such as \( \arcsin x, \arccos x, \arctan x \), are used to find angles given the values of trigonometric functions. Their derivatives are key in calculus, as they arise in integrals and differential equations. Derivatives for these functions are:

• \( \frac{d}{dx} \arcsin x = \frac{1}{\sqrt{1-x^2}} \)
• \( \frac{d}{dx} \arccos x = -\frac{1}{\sqrt{1-x^2}} \)
• \( \frac{d}{dx} \arctan x = \frac{1}{1+x^2} \)

These formulas allow you to differentiate expressions where inverse trigonometric functions are present. Pay close attention to signs and denominators, which often involve squares or square roots.