Problem 70 - Calculus for Biology and Medicine

Question

Differentiate with respect to the independent variable. $$ f(x)=\left(x^{3}-3 x^{2}+2\right)\left(\sqrt{x}+\frac{1}{\sqrt{x}}-1\right) $$

Step-by-Step Solution

Answer

The derivative of the function is: \[ f'(x) = 4x^{2.5} - 3x^{2} - 4x^{1.5} + 6x - 3x^{0.5} + 2 \frac{1}{x^{0.5}} \].

Step 1: Identify the Functions

Given the function \( f(x) = \left(x^3 - 3x^2 + 2\right)\left(\sqrt{x} + \frac{1}{\sqrt{x}} - 1\right) \), we notice it is a product of two separate functions, namely \( u(x) = x^3 - 3x^2 + 2 \) and \( v(x) = \sqrt{x} + \frac{1}{\sqrt{x}} - 1 \).

Step 2: Apply the Product Rule

To differentiate a product of two functions \( u(x) \) and \( v(x) \), we use the product rule: \( f'(x) = u'(x)v(x) + u(x)v'(x) \). We need to find the derivatives \( u'(x) \) and \( v'(x) \).

Step 3: Differentiate \( u(x) \)

Differentiate \( u(x) = x^3 - 3x^2 + 2 \):\[ u'(x) = 3x^2 - 6x \]This is found by applying the power rule to each term of \( u(x) \).

Step 4: Differentiate \( v(x) \)

Differentiate \( v(x) = \sqrt{x} + \frac{1}{\sqrt{x}} - 1 \). Rewriting \( v(x) \) using exponents: \( v(x) = x^{1/2} + x^{-1/2} - 1 \).Differentiate term by term:- \( \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{-1/2} \)- \( \frac{d}{dx}(x^{-1/2}) = -\frac{1}{2}x^{-3/2} \)- The derivative of \( -1 \) is 0.Thus, \( v'(x) = \frac{1}{2}x^{-1/2} - \frac{1}{2}x^{-3/2} \).

Step 5: Substitute into the Product Rule

Substitute \( u'(x) \), \( v(x) \), \( u(x) \), and \( v'(x) \) into the product rule formula:\[ f'(x) = (3x^2 - 6x)(\sqrt{x} + \frac{1}{\sqrt{x}} - 1) + (x^3 - 3x^2 + 2)\left(\frac{1}{2}x^{-1/2} - \frac{1}{2}x^{-3/2} \right) \]

Step 6: Simplify the Expression

Distribute and simplify:1. For \( (3x^2 - 6x)(\sqrt{x} + \frac{1}{\sqrt{x}} - 1) \): - Distribute \( 3x^2 \) and \( -6x \) through \( (\sqrt{x} + \frac{1}{\sqrt{x}} - 1) \).2. For \( (x^3 - 3x^2 + 2)\left(\frac{1}{2}x^{-1/2} - \frac{1}{2}x^{-3/2} \right) \): - Distribute \( x^3 \), \( -3x^2 \), and \( 2 \) through \( \frac{1}{2}x^{-1/2} - \frac{1}{2}x^{-3/2} \).3. Combine all like terms from both parts.After computation and simplification:\[ f'(x) = 3x^{2.5} + \frac{3}{2}x^{1.5} - 3x^{2} - x^{1.5} - 6x^{1.5} - 3x^{0.5} + 6x - \frac{3}{2}x^{0.5} + x^{2.5} - \frac{3}{2}x^{1.5} + x^{0.5} - \frac{1}{2}x^{1.5} + \frac{1}{2}x^{0.5} - 2x^{0.5} \].

Step 7: Provide the Final Simplified Answer

After further simplification of similar terms, the expression for the derivative is:\[ f'(x) = 4x^{2.5} - 3x^{2} - 4x^{1.5} + 6x - 3x^{0.5} + 2 \frac{1}{x^{0.5}} \].

Step 8: Verify the Answer

Double-check each derivative calculation and distribution for accuracy to ensure the final expression is correct.

Key Concepts

Product Rule for Differentiation

The product rule is a straightforward method for finding the derivative of a function that is the product of two differentiable functions. When you have a function like \( f(x) = u(x)v(x) \), the product rule states that the derivative \( f'(x) \) is given by the sum \( u'(x)v(x) + u(x)v'(x) \). This means you have to take the derivative of the first function and multiply it by the second, and vice versa, then add them together.

Let's break it down into steps to ease understanding:

• Identify the two functions being multiplied, say \( u(x) \) and \( v(x) \).
• Compute their derivatives, \( u'(x) \) and \( v'(x) \).
• Substitute these derivatives into the product rule formula: \( f'(x) = u'(x)v(x) + u(x)v'(x) \).

By following these steps, you can tackle more complex functions that involve products of variables. This rule simplifies what could otherwise be a cumbersome differentiation process, by allowing you to work with manageable pieces.

The Power Rule Explained

The power rule is a crucial tool in calculus differentiation, especially when dealing with polynomials. It tells us how to differentiate functions of the form \( x^n \). According to the power rule, if you have a term \( x^n \), its derivative is \( nx^{n-1} \). This rule simplifies the differentiation process of terms in a polynomial since you just need to apply it to each term.

Consider these steps:

• Take each term in the polynomial separately.
• Apply the power rule: multiply the exponent by the coefficient and decrease the exponent by one.
• Write down the derivative of each term.

In the exercise solution, functions like \( x^3 \) and \( x^{1/2} \) were differentiated using this rule, making it possible to find the derivative of each term in the expressions \( u(x) \) and \( v(x) \). This powerful rule streamlines the process and makes the task less daunting.

Simplifying Derivative Expressions

Derivative simplification is the process of combining and reducing terms to achieve a more manageable form. Once you have applied differentiation rules like the product and power rules, you often end up with a complicated expression that may contain similar terms.

Here's a guide to simplification:

• Distribute any common factors. In this case, resolve terms like \( (3x^2 - 6x)(\sqrt{x} + \frac{1}{\sqrt{x}} - 1) \).
• Combine like terms to reduce the number of terms in the expression.
• Reorganize terms based on their degree or power for clarity.
• Double-check each step to ensure that the simplification is accurate.

This process helps in presenting the final derivative in its neatest form, making the result much easier to interpret and work with. Through this simplification, the initially complex derivative expression becomes more concise, as demonstrated in the given solution.