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Find the derivative of each function by using the Product Rule. Simplify your answers. $$ f(x)=(\sqrt{x}-1)(\sqrt{x}+1) $$

Short Answer

Expert verified
The derivative of the function is 1.

Step by step solution

01

Identify the Functions

The given function is \[ f(x) = (\sqrt{x} - 1)(\sqrt{x} + 1) \]We will identify the first function as \( u = \sqrt{x} - 1 \) and the second function as \( v = \sqrt{x} + 1 \).
02

Find the Derivatives

Find the derivatives of both functions:- For \( u = \sqrt{x} - 1 \), the derivative \( u' = \frac{1}{2\sqrt{x}} \).- For \( v = \sqrt{x} + 1 \), the derivative \( v' = \frac{1}{2\sqrt{x}} \).
03

Apply the Product Rule

The Product Rule states that if \( f(x) = u(x)v(x) \), then the derivative \( f'(x) = u'v + uv' \).Applying this, we get:\[ f'(x) = \left(\frac{1}{2\sqrt{x}}\right)(\sqrt{x} + 1) + (\sqrt{x} - 1)\left(\frac{1}{2\sqrt{x}}\right) \]
04

Simplify the Expression

Distribute the derivatives in the expression:\[ f'(x) = \frac{\sqrt{x} + 1}{2\sqrt{x}} + \frac{\sqrt{x} - 1}{2\sqrt{x}} \]Combine like terms:\[ f'(x) = \frac{\sqrt{x} + 1 + \sqrt{x} - 1}{2\sqrt{x}} \]Simplify:\[ f'(x) = \frac{2\sqrt{x}}{2\sqrt{x}} = 1 \]
05

Verification

As a quick verification, observe that the function \( f(x) = (\sqrt{x} - 1)(\sqrt{x} + 1) = x - 1 \) when expanded.The derivative of \( x - 1 \) is indeed 1, confirming our solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivatives
When learning about derivatives, you're getting to grips with a fundamental concept in calculus. Derivatives tell us how a function changes at any given point, essentially giving us the function's rate of change or slope. This concept is crucial in various fields such as physics, engineering, and economics because it helps in understanding how quantities evolve over time or space.

To find the derivative of a function, we often use well-defined rules and formulas. For example, if you have a simple power function, the power rule can be applied. However, when you're dealing with products of functions, as in our exercise, the Product Rule comes into play. Knowing how to derive correctly is critical because it allows you to predict future behavior, optimize systems, and understand complex phenomena.
Simplifying Algebraic Expressions
Simplifying algebraic expressions means making them easier to work with by combining like terms or rewriting them in a simpler form. It often involves using basic algebraic techniques to reduce complex expressions into more manageable forms.

In the context of derivatives, simplifying your final answer can be extremely helpful. This step makes it easier to understand what the derivative represents and to work with further in applications.
  • Combine like terms: Check if there are common terms that you can add or subtract.
  • Factor where possible: If you can factor out common terms, do so to simplify further.
  • Cancel terms wisely: Be cautious when cancelling terms to avoid obvious mistakes.
Simplification ensures that your derivative is not only correct but also in its most understandable form, as seen in the given solution where the derivative was simplified to a neat '1'.
Chain Rule
Though not directly used in this exercise, understanding the Chain Rule is important for mastering derivatives, especially when dealing with composite functions. The Chain Rule allows you to differentiate a composite function, which is a function inside another function, and is defined by the formula \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \).

In scenarios where you encounter layers of functions, the Chain Rule becomes crucial. It breaks down the process into smaller and manageable chunks:
  • Differentiate the outer function while keeping the inner function unchanged.
  • Multiply by the derivative of the inner function.
Knowing when and how to apply the Chain Rule efficiently complements your skillset for tackling more complex calculus problems. With this powerful tool, even the most intricate functions become simpler to handle.

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Most popular questions from this chapter

True or False: If a function is not differentiable at a point, then its graph cannot have a tangent line at that point.

If a steel ball is tossed from the top of the Burj Khalifa in Dubai, the tallest building in the world, its height above the ground \(t\) seconds later will be \(s(t)=2717-16 t^{2}\) feet (neglecting air resistance). a. How long will it take to reach the ground? [Hint: Find when the height equals zero.] b. Use your answer to part (a) to find the velocity with which it will strike the ground. (This is called the impact velocity.) c. Find the acceleration at any time \(t\). (This number is called the acceleration due to gravity.)

For each piecewise linear function: a. Draw its graph (by hand or using a graphing calculator). b. Find the limits as \(x\) approaches 3 from the left and from the right. c. Is it continuous at \(x=3 ?\) If not, indicate the first of the three conditions in the definition of continuity (page 80 ) that is violated. \(f(x)=\left\\{\begin{array}{ll}5-x & \text { if } x \leq 3 \\ x-2 & \text { if } x>3\end{array}\right.\)

For each piecewise linear function: a. Draw its graph (by hand or using a graphing calculator). b. Find the limits as \(x\) approaches 3 from the left and from the right. c. Is it continuous at \(x=3 ?\) If not, indicate the first of the three conditions in the definition of continuity (page 80 ) that is violated. \(f(x)=\left\\{\begin{array}{ll}x & \text { if } x \leq 3 \\ 7-x & \text { if } x>3\end{array}\right.\)

Use the Quotient Rule to find a general expression for the marginal average revenue. That is, calculate \(\frac{d}{d x}\left[\frac{R(x)}{x}\right]\) and simplify your answer.

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