Problem 69 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The derivative is: \( f'(x) = -2\sqrt{2x} - \frac{4}{\sqrt{x}} + \frac{1}{\sqrt{2x}} - \frac{2x}{\sqrt{2x}} - \frac{1}{x^{3/2}} + \frac{2x}{x^{3/2}} \).

Step 1: Recognize the Product Rule

The function \( f(x) = (1-2x)\left(\sqrt{2x} + \frac{2}{\sqrt{x}}\right) \) is a product of two functions: \( u(x) = 1-2x \) and \( v(x) = \sqrt{2x} + \frac{2}{\sqrt{x}} \). Therefore, we will use the product rule: \( (uv)' = u'v + uv' \).

Step 2: Differentiate the First Function

Differentiate \( u(x) = 1-2x \) with respect to \( x \). The derivative is \( u'(x) = -2 \).

Step 3: Simplify and Differentiate the Second Function

Rewrite and differentiate \( v(x) = \sqrt{2x} + \frac{2}{\sqrt{x}} \). Simplify: \( v(x) = (2x)^{1/2} + 2x^{-1/2} \). Now differentiate: \( v'(x) = \frac{1}{2}(2x)^{-1/2} \times 2 + 2 \times (-\frac{1}{2}x^{-3/2}) = \frac{1}{\sqrt{2x}} - \frac{1}{x^{3/2}} \).

Step 4: Apply the Product Rule

Using the product rule: \( f'(x) = u'v + uv' \). Substitute \( u' = -2 \), \( u = 1-2x \), \( v = \sqrt{2x} + \frac{2}{\sqrt{x}} \), and \( v' = \frac{1}{\sqrt{2x}} - \frac{1}{x^{3/2}} \).

Step 5: Calculate Each Component of the Product Rule

Calculate \( -2 \times (\sqrt{2x} + \frac{2}{\sqrt{x}}) \) which is \( -2\sqrt{2x} - \frac{4}{\sqrt{x}} \). Now, calculate \((1-2x) \left(\frac{1}{\sqrt{2x}} - \frac{1}{x^{3/2}}\right)\).

Step 6: Combine and Simplify

Combine the results from Step 5: \( f'(x) = -2\sqrt{2x} - \frac{4}{\sqrt{x}} + \frac{1}{\sqrt{2x}} - \frac{2x}{\sqrt{2x}} - \frac{1}{x^{3/2}} + \frac{2x}{x^{3/2}} \). Simplify the expression to get the final derivative.

Key Concepts

Understanding the Product Rule

When tackling complex function differentiation, the product rule is a vital tool. It allows us to find the derivative of a product of two functions easily. If you have a function expressed as the product of two simpler functions, say \( u(x) \) and \( v(x) \), the product rule states that the derivative of this product \( f(x) = u(x)v(x) \) is given by:

• \( f'(x) = u'(x)v(x) + u(x)v'(x) \)

Breaking this down:

• Differentiate \( u(x) \) to get \( u'(x) \).
• Leave \( v(x) \) as it is for the first part of the calculation.
• Then, differentiate \( v(x) \) to get \( v'(x) \).
• Leave \( u(x) \) as it is for the second part.

These two differentiated products are then added together to find the derivative of the original function. It's important to correctly identify the two parts of the function and apply the rule accurately.

Exploring Derivatives

Derivatives are a fundamental part of calculus, representing the rate of change of a function with respect to a variable. If you imagine a curve, the derivative at any point gives you the slope of the tangent at that point. This concept helps in:

• Determining how fast a function is changing.
• Finding higher-dimensional analogs of slopes.
• Analyzing motion and other rate phenomena in physics and engineering.

For simple functions such as polynomials, finding derivatives typically involves power rules, such as reducing the exponent by one and multiplying by the original exponent. For more complicated functions, like those involving roots or fractions, it might involve using techniques such as the chain rule or product rule. In the provided exercise, understanding how to manipulate and differentiate functions like \( \sqrt{2x} \) and \( 2x^{-1/2} \) is crucial for successful differentiation.

Step-by-Step Differentiating Functions

Differentiating functions can seem daunting at first, but breaking down the process into clear, manageable steps helps immensely. Consider the function presented: \( f(x) = (1-2x)(\sqrt{2x} + \frac{2}{\sqrt{x}}) \). This consists of two main parts:

• First, identify each component and apply rules of differentiation.
• Use the product rule by differentiating the first function \( u(x) = 1-2x \), yielding \( u'(x) = -2 \).
• Next, rewrite \( v(x) \) as \( (2x)^{1/2} + 2x^{-1/2} \) to employ standard differentiation rules and obtain \( v'(x) = \frac{1}{\sqrt{2x}} - \frac{1}{x^{3/2}} \).

By systematically differentiating each function and plugging these derivatives into the product rule, you apply a technique to manage more complex problems. Always simplify the derivative expression to uncover the least complex form possible. Practice with different types of functions enhances familiarity and confidence in solving calculus problems.