Problem 33 - Essential Calculus Early Transcendentals

Question

If \(g(x)=x \sin ^{-1}(x / 4)+\sqrt{16-x^{2}},\) find \(g^{\prime}(2).\)

Step-by-Step Solution

Answer

\(g^{\prime}(2) = \frac{\pi}{6} - \frac{\sqrt{3}}{4} \).

Step 1: Substitute and Differentiate

The function is given as \( g(x) = x \sin^{-1}\left( \frac{x}{4} \right) + \sqrt{16-x^2} \). To find \( g'(x) \), apply the product rule to differentiate the first term and the chain rule to differentiate the second term.

Step 2: Differentiate First Term

The first term is \( x \sin^{-1}\left( \frac{x}{4} \right) \). The derivative can be found using the product rule:\[ \left( u \cdot v \right)' = u'v + uv' \] where \( u = x \) and \( v = \sin^{-1}\left( \frac{x}{4} \right) \). So, \( u' = 1 \) and \[ v' = \frac{1}{\sqrt{1-\left( \frac{x}{4} \right)^2}} \cdot \frac{1}{4} = \frac{1}{4\sqrt{16-x^2}} \]. Therefore, the derivative of the first term is: \[ 1 \sin^{-1}\left( \frac{x}{4} \right) + x \cdot \frac{1}{4\sqrt{16-x^2}} \].

Step 3: Differentiate Second Term

The second term is \( \sqrt{16-x^2} \). Using the chain rule, the derivative is:\[ \frac{d}{dx} \left( \sqrt{16-x^2} \right) = \frac{1}{2\sqrt{16-x^2}} \cdot (-2x) = -\frac{x}{\sqrt{16-x^2}} \].

Step 4: Combine Results

Combine the derivatives obtained from Steps 2 and 3 to find \( g'(x) \):\[ g'(x) = \sin^{-1}\left( \frac{x}{4} \right) + \frac{x}{4\sqrt{16-x^2}} - \frac{x}{\sqrt{16-x^2}} \]. Simplify the expression, combining terms inside the square roots:\[ g'(x) = \sin^{-1}\left( \frac{x}{4} \right) - \frac{3x}{4\sqrt{16-x^2}} \].

Step 5: Evaluate at x = 2

Substitute \( x = 2 \) into \( g'(x) \):\[ g'(2) = \sin^{-1}\left( \frac{2}{4} \right) - \frac{3 \times 2}{4 \times \sqrt{16-2^2}} \].Firstly, calculate \( \sin^{-1}\left( \frac{1}{2} \right) \), which corresponds to \( \frac{\pi}{6} \) radians. Then evaluate \( \sqrt{16-4} = \sqrt{12} = 2\sqrt{3} \).Finally, we have:\[ g'(2) = \frac{\pi}{6} - \frac{6}{8\sqrt{3}} = \frac{\pi}{6} - \frac{3}{4\sqrt{3}} \]. Rationalize the denominator of the second fraction by multiplying the numerator and the denominator by \( \sqrt{3} \):\[ g'(2) = \frac{\pi}{6} - \frac{3\sqrt{3}}{12} = \frac{\pi}{6} - \frac{\sqrt{3}}{4} \].

Key Concepts

Product Rule

The product rule in calculus is a crucial tool for finding derivatives when faced with the product of two differentiable functions. Essentially, if you have two functions, say \( u(x) \) and \( v(x) \), and you want to differentiate their product \( u(x) \cdot v(x) \), the product rule states that:

• The derivative of \( u(x)v(x) \) is \( u'(x)v(x) + u(x)v'(x) \).
• This means you take the derivative of the first function multiplied by the second, plus the first function multiplied by the derivative of the second.

This rule is particularly useful when tackling complex expressions where simple term-by-term differentiation isn’t feasible. In our example, the function \( g(x) = x \sin^{-1}(x/4) + \sqrt{16-x^2} \) involves the product of \( x \) and \( \sin^{-1}(x/4) \), necessitating the use of the product rule for proper differentiation. This step-by-step breakdown aids in systematically finding the derivative without missing any necessary components.

Chain Rule

The chain rule is a powerful technique for finding the derivative of composite functions, which are functions nested within other functions. When you have a function \( f(g(x)) \), the chain rule helps to differentiate it in terms of its inner and outer functions. The formula is:

• Derivative of \( f(g(x)) \) is \( f'(g(x)) \cdot g'(x) \).
• This means you differentiate the outer function at the inner function, then multiply by the derivative of the inner function.

Consider the second term in the function \( g(x) = x \sin^{-1}(x/4) + \sqrt{16-x^2} \). Here, \( \sqrt{16-x^2} \) is a composite function, consisting of an inner function \( 16-x^2 \) and an outer function \( \sqrt{\cdot} \). The chain rule facilitates the differentiation of \( \sqrt{16-x^2} \) by simplifying this complex structure into manageable steps.

Inverse Trigonometric Functions

Inverse trigonometric functions are essential components in calculus, especially when you need to solve problems involving angles derived from trigonometric ratios. They reverse the process of the standard trigonometric functions. A key challenge is differentiating these functions, particularly \( \sin^{-1}(x) \), \( \cos^{-1}(x) \), and \( \tan^{-1}(x) \).

• The derivative of \( \sin^{-1}(x) \) is \( \frac{1}{\sqrt{1-x^2}} \).
• In complex expressions, such as \( \sin^{-1}(x/4) \), the differentiation often forms a part of a larger process, such as with the product rule.

Using these derivatives is essential in accurately calculating the rate of change, particularly when the function involves inverse operations, just as our given example function does. Careful attention to these derivatives ensures precision in solving calculus problems.

Derivative of Square Root Function

The derivative of a square root function is a key concept when dealing with functions involving square roots. It involves using both the power rule and the chain rule for differentiation. For a function of the form \( \sqrt{h(x)} \), the derivative is calculated as:

• Convert the square root to a power expression: \( h(x)^{1/2} \).
• The derivative is \( \frac{1}{2}h(x)^{-1/2}h'(x) \), which simplifies to \( \frac{h'(x)}{2\sqrt{h(x)}} \).

In the function \( g(x) = x \sin^{-1}(x/4) + \sqrt{16-x^2} \), the second term \( \sqrt{16-x^2} \) is differentiated using this approach. Using the chain rule, we apply this methodical approach to find its derivative. This strategy helps in dealing with non-linear functions and captures the profound relationship between functions and their rates of change.