Problem 27 - Applied Calculus

Question

Find the derivative. Assume \(a, b, c, k\) are constants. $$y=\sqrt{x}(x+1)$$

Step-by-Step Solution

Answer

The derivative is \( y' = \frac{3}{2}x^{1/2} + \frac{1}{2x^{1/2}} \).

Step 1: Rewrite the Expression

The given function is \( y = \sqrt{x}(x+1) \). Start by rewriting \( \sqrt{x} \) as \( x^{1/2} \). Thus, the function becomes \( y = x^{1/2}(x+1) \).

Step 2: Identify and Apply Product Rule

We have a product of two functions: \( u = x^{1/2} \) and \( v = x + 1 \). The derivative of a product \( uv \) is given by \((uv)' = u'v + uv' \). We need to find \( u' \) and \( v' \).

Step 3: Differentiate Each Function

Differentiate \( u = x^{1/2} \) to find \( u' = \frac{1}{2}x^{-1/2} \). Differentiate \( v = x + 1 \) to find \( v' = 1 \).

Step 4: Apply the Product Rule

Substitute \( u' \), \( u \), \( v' \), and \( v \) into the product rule formula: \[ (uv)' = \frac{1}{2}x^{-1/2}(x+1) + x^{1/2}(1) \].

Step 5: Simplify the Derivative Expression

Simplify each term: - The first term is \( \frac{1}{2}x^{-1/2}(x+1) = \frac{1}{2}x^{-1/2}x + \frac{1}{2}x^{-1/2} = \frac{1}{2}x^{1/2} + \frac{1}{2x^{1/2}} \).- The second term is \( x^{1/2} \).Combine them to get: \[ y' = \frac{1}{2}x^{1/2} + \frac{1}{2x^{1/2}} + x^{1/2} \].

Step 6: Combine Like Terms

Combine like terms: the two terms with \( x^{1/2} \) are \( \frac{1}{2}x^{1/2} \) and \( x^{1/2} \). Thus, \( \frac{1}{2}x^{1/2} + x^{1/2} = \frac{3}{2}x^{1/2} \).Finally, the derivative is:\[ y' = \frac{3}{2}x^{1/2} + \frac{1}{2x^{1/2}} \].

Key Concepts

Product Rule

In calculus, when you have a function that is the product of two or more simpler functions, you can't just take the derivative of each part separately. Instead, you use the product rule.

• The product rule states that the derivative of a product of two functions is given by: \[ (uv)' = u'v + uv' \]

This formula means you take the derivative of the first function, multiply it by the second function, and then add the product of the first function and the derivative of the second function.
This approach ensures that both components contribute to the rate of change of the whole product function over time. In our exercise, the function \( y = \sqrt{x}(x+1) \) is composed of two parts:

• \( u = x^{1/2} \)
• \( v = x + 1 \)

By using the product rule, we systematically calculated the contributions from both functions to find the overall derivative. This method is fundamental, especially when dealing with more complex functions in calculus.

Differentiation

Differentiation is one of the core operations in calculus. It involves finding the derivative of a function, which is essentially the rate at which the function's value is changing at any given point. When differentiating, you're looking for how the outputs of a function change with respect to its inputs.For the function \( y = \sqrt{x}(x+1) \), we first rewrote \( \sqrt{x} \) as \( x^{1/2} \) to make differentiation easier. Then, we found the derivatives of the individual components of the function:

• \( u = x^{1/2} \) differentiates to \( u' = \frac{1}{2}x^{-1/2} \)
• \( v = x + 1 \) differentiates to \( v' = 1 \)

Identifying these derivatives allowed us to apply the product rule effectively. Differentiation involves various rules, such as the power rule, the product rule, and the chain rule. In this exercise, combining these rules is crucial to accurately finding how our original function changes.

Simplifying Expressions

Simplifying expressions is a key step in ensuring that derivatives and other mathematical results are clear and straightforward. After applying the product rule and obtaining the initial form of the derivative, the next step is to simplify it.During simplification, we often look to:

• Combine like terms to reduce the complexity.
• Reorganize the expression for a simplified look that is easier to interpret.

In this example, after using the product rule, the derivative was: \[ y' = \frac{1}{2}x^{-1/2}(x+1) + x^{1/2}(1) \].This expanded form needs simplification. Breaking it down:- The first term \( \frac{1}{2}x^{-1/2}(x+1) \) was expanded and combined as \( \frac{1}{2}x^{1/2} + \frac{1}{2x^{1/2}} \).- Different parts of the expression with \( x^{1/2} \) were combined: \( \frac{1}{2}x^{1/2} + x^{1/2} = \frac{3}{2}x^{1/2} \).Through simplifying, the final derivative expression became clearer and more concise: \[ y' = \frac{3}{2}x^{1/2} + \frac{1}{2x^{1/2}} \].