Problem 59 - Calculus Early Transcendentals

Question

(a) Use a graphing utility to obtain the graph of the function \(f(x)=x \sqrt{4-x^{2}}\) (b) Use the graph in part (a) to make a rough sketch of the graph of \(f^{\prime}\). (c) Find \(f^{\prime}(x),\) and then check your work in part (b) by using the graphing utility to obtain the graph of \(f^{\prime} .\) (d) Find the equation of the tangent line to the graph of \(f\) at \(x=1,\) and graph \(f\) and the tangent line together.

Step-by-Step Solution

Answer

Compute the derivative \( f'(x) = \frac{4-2x^2}{\sqrt{4-x^2}} \), and verify graphically; the tangent at \( x=1 \) is \( y = \frac{2}{\sqrt{3}}x - \frac{2}{\sqrt{3}} + \sqrt{3} \).

Step 1: Graph the Function

First, input the function \( f(x) = x \sqrt{4-x^2} \) into a graphing utility. This might be a graphing calculator or an online tool. Observe the shape and key features of the graph (such as intercepts, maximum points, and symmetry). The domain of this function is \(-2 \leq x \leq 2\) because the expression inside the square root must be non-negative.

Step 2: Sketch the Derivative Graph

Based on the graph from Step 1, make a rough sketch of what the derivative \( f'(x) \) might look like. Look for slopes of tangent lines at various points along the curve; these slopes are represented by the height of the \( f'(x) \) graph. Where \( f(x) \) has a maximum or minimum, \( f'(x) \) should cross the x-axis.

Step 3: Compute the Derivative

To find \( f'(x) \), apply the product rule and chain rule. The function can be expressed as \( f(x) = x \cdot (4-x^2)^{1/2} \). Using differentiation rules:\[ f'(x) = \frac{d}{dx} \left( x \right) \cdot \sqrt{4-x^2} + x \cdot \frac{d}{dx} \left( \sqrt{4-x^2} \right) \]This expands to:\[ f'(x) = \sqrt{4-x^2} + x \cdot \left( -\frac{x}{\sqrt{4-x^2}} \right) \]Simplify to:\[ f'(x) = \frac{4-x^2-x^2}{\sqrt{4-x^2}} = \frac{4-2x^2}{\sqrt{4-x^2}} \]

Step 4: Verify the Derivative Graphically

Use the graphing utility to graph \( f'(x) = \frac{4-2x^2}{\sqrt{4-x^2}} \) and compare it with your sketch from Step 2. Ensure that features such as where the derivative crosses the x-axis and regions of increase/decrease match what was observed and sketched.

Step 5: Find the Tangent Line Equation

At \( x = 1 \), calculate \( f(1) = 1 \times \sqrt{4-1^2} = \sqrt{3} \) and \( f'(1) = \frac{4-2(1)^2}{\sqrt{4-(1)^2}} = \frac{2}{\sqrt{3}} \). The equation of the tangent line in point-slope form is given by:\[ y - \sqrt{3} = \frac{2}{\sqrt{3}}(x-1) \]Simplify to find:\[ y = \frac{2}{\sqrt{3}}x - \frac{2}{\sqrt{3}}(1) + \sqrt{3} \]

Step 6: Graph Function and Tangent Line

Plot \( f(x) = x\sqrt{4-x^2} \) and the tangent line \( y = \frac{2}{\sqrt{3}}x - \frac{2}{\sqrt{3}} + \sqrt{3} \) together on the same graphing utility. Confirm the tangent line just touches the curve at the point \( (1, \sqrt{3}) \) without crossing it.

Key Concepts

Tangent Line

Understanding tangent lines is an important part of calculus. Imagine placing a ruler against a curve; if it touches the curve at just one point without cutting through, it acts like a tangent line. A tangent line to the graph of a function at a point basically gives us the direction in which the curve is heading at that specific point. This means that the slope of the tangent line is equal to the derivative of the function at that point.

Here's how you figure out the equation of a tangent line: you need two things—the slope of the tangent (which is the derivative) and the point where the line touches the curve. Once you have these, you can use the point-slope form of a line:

• First, find the derivative of the function.
• Evaluate this derivative at the given x-value to get the slope.
• Plug this slope and the point into the point-slope formula: \[y - y_1 = m(x - x_1)\]where \(\, (x_1, y_1)\) is the point and\(\, m\) is the slope.

Use this information to draw the tangent line on your graph. In our example, the tangent line to the function \(f(x) = x\sqrt{4-x^2}\) at \(x=1\) would use the calculated slope and point to help illustrate how the curve behaves right there.

Product Rule

When working with functions that are products of two other functions, such as \(f(x) = x \cdot (4-x^2)^{1/2}\), you need the product rule for differentiation. This rule is extremely useful, and it goes like this: if you have a function \(u(x)\) times another function \(v(x)\), then the derivative \((uv)'\) is:

• \[(uv)' = u'v + uv'\]

This means that you need to find the derivative of the first function, multiply it by the second function as it is, and then add the product of the first function as it is with the derivative of the second function.

In the original exercise, using the product rule lets us handle the function \(x\sqrt{4-x^2}\) easily. We differentiate \(x\) (which becomes 1) and multiply by \((4-x^2)^{1/2}\), then take \(x\) and multiply it by the derivative of \((4-x^2)^{1/2}\) obtained using the chain rule. It's a straightforward way to deal with these kinds of expressions, making it an essential tool in your calculus toolkit.

Chain Rule

The chain rule is a fundamental concept for differentiating composite functions—functions within another function. If you have a function in the form \(f(g(x))\), the chain rule helps you find the derivative by taking the derivative of the outer function and multiplying it by the derivative of the inner function.

• The formula is: \[(f(g(x)))' = f'(g(x)) \cdot g'(x)\]

Applying the chain rule requires knowing how to recognize when functions are "nested" together and carefully applying this order of operations.

In the exercise at hand, the chain rule comes into play when differentiating \((4-x^2)^{1/2}\). Here, the outer function is the square root operation, and the inner function is \(4-x^2\). Applying the chain rule allows us to find the derivative of the square root function first, treating the inner function as a single variable, and then differentiate the inside function separately, as demonstrated when calculating \(f'(x)\). Being comfortable with the chain rule makes tackling a wide range of problems much easier.