Problem 21 - Essential Calculus Early Transcendentals

Question

Find the derivative of the function. Simplify where possible. $$G(x)=\sqrt{1-x^{2}} \text { arccos } x$$

Step-by-Step Solution

Answer

The derivative of \( G(x) = \sqrt{1-x^2} \arccos(x) \) is \( G'(x) = \frac{-x \arccos(x)}{\sqrt{1-x^2}} - 1 \).

Step 1: Identify the Function Components

The function to differentiate is given by \( G(x) = \sqrt{1-x^2} \arccos(x) \). This function is a product of two functions: \( u(x) = \sqrt{1-x^2} \) and \( v(x) = \arccos(x) \). We will use the product rule for differentiation.

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

The function \( u(x) = \sqrt{1-x^2} \) can be rewritten as \( (1-x^2)^{1/2} \). To find its derivative, use the chain rule. The derivative is:\[ u'(x) = \frac{1}{2}(1-x^2)^{-1/2}(-2x) = \frac{-x}{\sqrt{1-x^2}} \]

Step 3: Differentiate \( v(x) = \arccos(x) \)

The derivative of \( v(x) = \arccos(x) \) is a known formula:\[ v'(x) = \frac{-1}{\sqrt{1-x^2}} \]

Step 4: Apply the Product Rule

The product rule states that the derivative of a product \( u(x)v(x) \) is \( u'(x)v(x) + u(x)v'(x) \). Applying this rule, we have:\[ G'(x) = u'(x)v(x) + u(x)v'(x) \]

Step 5: Substitute the Derivatives and Simplify

Substitute the derivatives derived earlier into the product rule formula:\[ G'(x) = \left(\frac{-x}{\sqrt{1-x^2}}\right) \arccos(x) + \sqrt{1-x^2}\left(\frac{-1}{\sqrt{1-x^2}}\right) \]Simplify the expression:\[ G'(x) = \frac{-x \arccos(x)}{\sqrt{1-x^2}} - 1 \]

Key Concepts

Product Rule Differentiation

When working with derivatives, the product rule is a fundamental tool to have in your mathematics toolkit. It helps us find the derivative of a function that is the product of two other functions. Here's how the product rule works: if you have two functions, say \( u(x) \) and \( v(x) \), and you need to differentiate their product \( u(x)v(x) \), the product rule states:

• The derivative \( (uv)' = u'v + uv' \)

Break this down to understand it better:- Differentiate the first function (denoted as \( u' \)), then multiply it by the second function as it is.

Add to that: the first function as it is, multiplied by the derivative of the second function (denoted as \( v' \)).

In our example, \( G(x) = \sqrt{1-x^2} \arccos(x) \), we apply the product rule after identifying that \( u(x) = \sqrt{1-x^2} \) and \( v(x) = \arccos(x) \). This leads to using the derivatives \( u'(x) \) and \( v'(x) \) obtained from the further steps to calculate \( G'(x) \). Look back at step 5 of the solution for the final simplified form.

Chain Rule in Calculus

The chain rule is a handy rule for finding the derivative of compositions of functions. When you have a function inside another function, the chain rule lets you differentiate them together.Consider a composite function \( u(x) = (1-x^2)^{1/2} \). Here, you can see the square root function and the polynomial \( 1-x^2 \), which are chained together. The chain rule formula is:

• If you have \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x))g'(x) \)

Let's apply this to differentiate \( u(x) \):- First, identify the outer function \( f(z) = z^{1/2} \) and the inner function \( g(x) = 1-x^2 \).

Differentiate the outer function: \( f'(z) = \frac{1}{2}z^{-1/2} \)

Differentiate the inner function: \( g'(x) = -2x \)

Multiply these derivatives, substituting back the inner function, to get:\( u'(x) = \frac{1}{2}(1-x^2)^{-1/2}(-2x) = \frac{-x}{\sqrt{1-x^2}} \).This is an example of the power of the chain rule, allowing you to tackle more complex derivatives.

Derivatives of Inverse Trigonometric Functions

Inverse trigonometric functions, like \( \arccos(x) \), play an important role in calculus. Their derivatives have formulas that you need to memorize.For the arccosine function, the derivative formula is:

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

Why does it look like that? The derivative of \( \arccos(x) \) stems from its relationship with the cosine function. Think about how angles work within these trigonometric contexts – the derivative reflects this inverse relationship.In the original solution, we quickly identified \( v(x) = \arccos(x) \) and used the known derivative result. Applying these formulas saves time and avoids mistakes. Keep these derivatives in mind, especially when you work with products or compositions involving trigonometric functions.Use these concepts to easily differentiate products of functions and solve complex calculus problems with confidence.