Problem 70 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

Use the product rule to find the derivative of each part, then simplify the resulting expression.

Step 1: Recognize the Product Rule

The function given is the product of two separate functions: \(u(x) = x^2 - 1\) and \(v(x) = \sqrt{x} + \frac{1}{\sqrt{x}} - 1\). To differentiate this, we need to use the product rule, which states if you have \(f(x) = u(x) \cdot v(x)\), then \(f'(x) = u'(x) v(x) + u(x) v'(x)\).

Step 2: Differentiate \(u(x) = x^2 - 1\)

Differentiate \(u(x)\) with respect to \(x\):\[u'(x) = \frac{d}{dx}(x^2 - 1) = 2x.\]

Step 3: Simplify \(v(x) = \sqrt{x} + \frac{1}{\sqrt{x}} - 1\)

First, express \(v(x)\) in terms of powers of \(x\):\[v(x) = x^{1/2} + x^{-1/2} - 1.\]

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

Differentiate \(v(x)\) with respect to \(x\):\[v'(x) = \frac{d}{dx}(x^{1/2} + x^{-1/2} - 1) = \frac{1}{2}x^{-1/2} - \frac{1}{2}x^{-3/2}.\]

Step 5: Apply the Product Rule

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

Step 6: Simplify the Expression

Expand and simplify the expression from Step 5:\[f'(x) = 2x \cdot (\sqrt{x} + \frac{1}{\sqrt{x}} - 1) + (x^2 - 1)(\frac{1}{2}x^{-1/2} - \frac{1}{2}x^{-3/2}).\]Calculate each part separately if needed, and combine the terms to simplify the derivative.

Key Concepts

Product Rule

The product rule is a fundamental concept in calculus used to differentiate products of two functions. When dealing with a function defined as the product of two separate functions, such as \( f(x) = u(x) \cdot v(x) \), the product rule comes into play.
The formula for the product rule is simple:

• Differentiate the first function, \( u(x) \), while keeping the second function, \( v(x) \), unchanged.
• Then, differentiate the second function while keeping the first function unchanged.
• Finally, add the results from both differentiations. This gives you \( f'(x) = u'(x) v(x) + u(x) v'(x) \).

This rule is especially useful because it allows us to break down complex functions into simpler parts, enabling easier differentiation. In the exercise, the product rule helped us manage the differentiation of \((x^2 - 1)\) and \((\sqrt{x} + \frac{1}{\sqrt{x}} - 1)\) systematically.

Chain Rule

The chain rule is another essential tool in differentiation, often used when dealing with compositions of functions. Although the original exercise does not explicitly use the chain rule, understanding it deepens your comprehension of differentiating complex functions.
Imagine you have a composite function like \( y = f(g(x)) \). The chain rule allows you to differentiate this by focusing on the outer and inner functions separately:

• First, differentiate the outer function with respect to the inner function, which is \( f'(g(x)) \).
• Then, multiply this by the derivative of the inner function, \( g'(x) \).
• The result is: \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).

In practice, the chain rule is indispensable when functions are nested within each other, making it possible to break down the differentiation process into manageable parts.

Power Functions

Power functions, such as \(x^n\), where \(n\) is any real number, have a straightforward differentiation process. They are a common element in calculus and often appear in various forms.
Every power function can be differentiated using the power rule, which states:

• If \(f(x) = x^n\), then its derivative, \(f'(x)\), is \(nx^{n-1}\).

This rule is vital because it applies to all power functions, including those with fractional or negative exponents. In our exercise, we used this rule to differentiate \(x^{1/2}\) and \(x^{-1/2}\), which are just examples of power functions with fractional exponents. Understanding how to differentiate power functions lays a solid foundation for tackling more complex functions.