Problem 99 - Calculus and its applications

Question

Find the derivative of each of the following functions. Then use a calculator to check the results. $$ f(x)=x \sqrt{4-x^{2}} $$

Step-by-Step Solution

Answer

The derivative of \(f(x) = x \sqrt{4-x^2}\) is \(f'(x) = \frac{4-2x^2}{\sqrt{4-x^2}}\).

Step 1: Recognize the Product Rule

The given function is a product of two functions, \(f(x) = x \cdot \sqrt{4-x^2}\). When differentiating, we'll use the product rule, which states: \((uv)' = u'v + uv'\) where \(u = x\) and \(v = \sqrt{4-x^2}\).

Step 2: Differentiate the First Function

First, differentiate \(u = x\). The derivative is straightforward: \(u' = 1\).

Step 3: Differentiate the Second Function Using the Chain Rule

For \(v = \sqrt{4-x^2}\), rewrite it as \(v = (4-x^2)^{1/2}\). To differentiate, use the chain rule. First, the derivative of \((4-x^2)^{1/2}\) is \(\frac{1}{2}(4-x^2)^{-1/2}\). Then, multiply by the derivative of the inside function, \(-2x\). Thus, \(v' = \frac{1}{2}(4-x^2)^{-1/2} \cdot (-2x) = -\frac{x}{\sqrt{4-x^2}}\).

Step 4: Apply the Product Rule

Use the derivatives found in the previous steps in the product rule formula: \((uv)' = u'v + uv'\). Substituting the derivatives, we get:\(f'(x) = 1 \cdot \sqrt{4-x^2} + x \cdot \left(-\frac{x}{\sqrt{4-x^2}}\right)\).

Step 5: Simplify the Result

Simplify the expression for the derivative:\(f'(x) = \sqrt{4-x^2} - \frac{x^2}{\sqrt{4-x^2}}\). Combine the terms over a common denominator: \(f'(x) = \frac{(4-x^2)-x^2}{\sqrt{4-x^2}} = \frac{4-2x^2}{\sqrt{4-x^2}}\).

Step 6: Use a Calculator to Check the Result

To verify the result using a calculator, specify additional points to check the derivative values. Directly input the original function, differentiate it, and evaluate at different values of \(x\). Ensure calculated values match simplified expressions: \(f'(x) = \frac{4-2x^2}{\sqrt{4-x^2}}\) should be consistent with the numerical result.

Key Concepts

Product Rule

When you're asked to find the derivative of a function like \( f(x) = x \sqrt{4-x^2} \), you are actually dealing with a product of two separate functions. This is where the product rule in calculus comes into play. The product rule is quite straightforward and can be a nifty tool in your calculus toolkit.

In essence, the product rule states that if you have two functions \( u \) and \( v \), their product \( (uv) \) has a derivative that is calculated as \((uv)' = u'v + uv'\). In our exercise, we identified \( u = x \) and \( v = \sqrt{4-x^2} \).

Here’s how it works:

• First, differentiate \( u = x \) to get \( u' = 1 \).
• Next, move on to \( v = \sqrt{4-x^2} \), which will require the chain rule. More on that soon!
• Combine these results using the product rule to find the derivative of the entire function.

This allows us to break down complex functions into manageable pieces by focusing on one part at a time.

Chain Rule

The chain rule is an essential concept in calculus when you need to differentiate composite functions. A composite function is a function inside another function. It's like layers of an onion, and the chain rule helps you peel them one by one.

In the case of our function \( v = \sqrt{4-x^2} \), rewrite this square root as \( (4-x^2)^{1/2} \) to differentiate it more easily. Notice that it's not just any ordinary function, but an operation involving the power and the inside function \( 4-x^2 \).

• First, differentiate the outer function \((4-x^2)^{1/2}\), resulting in \(\frac{1}{2}(4-x^2)^{-1/2}\).
• Then, multiply this by the derivative of the inside function \(-2x\).

This step gives the inner workings of how differentiating composite functions works and results in: \[v' = -\frac{x}{\sqrt{4-x^2}} \]By using the chain rule, you ensure you don't miss any part of this layered function structure.

Function Differentiation

Differentiating functions is at the heart of calculus. It allows you to determine how fast or slow something is changing. This process involves recognizing various function forms and applying the right rules, like the product rule and chain rule.

With our function \( f(x) = x \sqrt{4-x^2} \), differentiation combines both these rules. Start with identifying these separate components or functions within your main function:

• \( u = x \) is simple and straightforward, providing us with \( u' = 1 \).
• \( v = \sqrt{4-x^2} \) is more complex, needing the chain rule to differentiate.

Once both components are differentiated, you can assemble them into a single expression using the product rule: \[(uv)'= u'v + uv' = 1 \cdot \sqrt{4-x^2} + x \cdot \left(-\frac{x}{\sqrt{4-x^2}}\right)\]This will lead to simplifying what you get into an elegant final derivative: \[f'(x) = \frac{4-2x^2}{\sqrt{4-x^2}}\]Function differentiation offers insight into the subtleties of change, combining various differentiation techniques for smooth calculations.