Problem 62 - Calculus for Biology and Medicine

Question

Differentiate with respect to the independent variable. \(f(x)=\frac{\sqrt{5 x}\left(1+x^{2}\right)}{\sqrt{2}}\)

Step-by-Step Solution

Answer

Use the quotient rule and simplify the terms to differentiate the function.

Step 1: Understand the Problem

We need to differentiate the function \(f(x) = \frac{\sqrt{5x}(1 + x^2)}{\sqrt{2}}\) with respect to \(x\). This function involves a quotient, so we will use the quotient rule of differentiation.

Step 2: Apply the Quotient Rule

The quotient rule states that if you have a function \(\frac{u}{v}\), its derivative is \(\frac{v\,du/dx - u\,dv/dx}{v^2}\). Here, \(u = \sqrt{5x}(1 + x^2)\) and \(v = \sqrt{2}\). Since \(v\) is a constant, \(dv/dx = 0\), which simplifies our problem.

Step 3: Find \(du/dx\) for the Numerator

First, differentiate \(u = \sqrt{5x}(1 + x^2)\) using the product rule since it is a product of \(\sqrt{5x}\) and \((1 + x^2)\). The product rule states \(d(uv)/dx = u \cdot dv/dx + v \cdot du/dx\).

Step 4: Differentiate \(\sqrt{5x}\) and \(1 + x^2\)

For \(\sqrt{5x} = (5x)^{1/2}\), the derivative is \(\frac{1}{2}(5x)^{-1/2} \cdot 5 = \frac{5}{2\sqrt{5x}}\). For \(1 + x^2\), the derivative is \(2x\).

Step 5: Calculate \(du/dx\) Using the Product Rule

Using the product rule: \(du/dx = (\sqrt{5x}) \cdot (2x) + (1 + x^2) \cdot \frac{5}{2\sqrt{5x}}\). Simplifying this gives us \(2x\sqrt{5x} + \frac{5(1+x^2)}{2\sqrt{5x}}\).

Step 6: Put it All Together

Our original function was \(f(x) = \frac{u}{v} = \frac{\sqrt{5x}(1 + x^2)}{\sqrt{2}}\). We now apply the quotient rule: \(f'(x) = \frac{\sqrt{2} \cdot du/dx)}{2}\). Substitute \(du/dx\) to get \(f'(x) = \frac{\sqrt{2}(2x\sqrt{5x} + \frac{5(1+x^2)}{2\sqrt{5x}})}{2}\). Simplify further if necessary for the final derivative.

Key Concepts

Quotient Rule

When you have a function that is a fraction, like in the exercise, the quotient rule is the go-to method for differentiation. This rule helps us handle functions that involve division. Imagine your function as a fraction where the top part is called the "numerator" and the bottom part is the "denominator."The quotient rule formula is:\[ \frac{d}{dx}\left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]Here:

• \(u\) is the numerator, the top part of the fraction.
• \(v\) is the denominator, the bottom part.

In the given problem:

• \(u = \sqrt{5x}(1 + x^2)\)
• \(v = \sqrt{2}\)

Since \(v\) is a constant, only \(u\) contributes to the differentiation. The process becomes a bit simpler when the denominator holds a constant value, because its derivative, \(dv/dx\), will be zero.

Product Rule

The product rule assists in finding the derivative of two functions that are multiplied together. In our exercise, the term \(\sqrt{5x}(1 + x^2)\) is a product of two functions: \(\sqrt{5x}\) and \(1 + x^2\).The product rule formula is:\[ \frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx} \]Breaking it further:

• \(u\) - The first function to differentiate, which is \(\sqrt{5x}\).
• \(v\) - The second function, \(1 + x^2\).

This means:

• You treat \(\sqrt{5x}\) as constant when differentiating \(1 + x^2\), and vice versa.

This interchanging differentiation simplifies finding derivatives in product-form functions.

Derivative of Square Root Functions

Square roots may look a bit intimidating, but they're quite manageable. The derivative of a square root, like \(\sqrt{5x}\), involves converting it into an exponent form.Remember:\[ \sqrt{5x} = (5x)^{1/2} \]To differentiate it, we use the power rule:

• Bring down the exponent as a factor.
• Subtract one from the exponent.

For \((5x)^{1/2}\), it becomes:

• \(\frac{1}{2}(5x)^{-1/2} \cdot 5 = \frac{5}{2\sqrt{5x}} \)

Notice how multiplying by the derivative of \(5x\) affects the final result. Applying this method streamlines differentiating any square root function.

Chain Rule

The chain rule is invaluable when dealing with composite functions—functions within functions. While our main exercise here doesn't prominently use the chain rule, understanding it is beneficial, especially when square roots or other more complex functions need differentiation.The chain rule formula is:\[ \frac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x) \]Simply put:

• Differentiate the outer function, leave the inner function inside.
• Multiply by the derivative of the inner function.

Even though the square root in our main exercise didn't directly require the chain rule, it's a vital tool when dissecting nested functions. Always ensure to adjust for any inner derivative when you're peeling back layers of a more complex expression.