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

The derivative is \( f'(x) = \frac{15x^2 + 5}{2\sqrt{10x^3 + 10x}}. \)

Step 1: Setup the Function for Differentiation

The function given is \( f(x) = \frac{\sqrt{5x(1+x^2)}}{\sqrt{2}} \). This can be rewritten as \( f(x) = \frac{1}{\sqrt{2}} \cdot \sqrt{5x + 5x^3} \). For differentiation, we need to use the chain rule and, potentially, the product rule.

Step 2: Differentiate Using the Chain Rule

The function can be modeled as \( u(x) = \sqrt{5x^3 + 5x} \) where \( f(x) = \frac{1}{\sqrt{2}} u(x) \). The derivative \( u'(x) \) uses the chain rule: \( \frac{1}{2\sqrt{5x^3 + 5x}} \cdot (15x^2 + 5) \).

Step 3: Apply the Constant Multiple Rule

As \( f(x) = \frac{1}{\sqrt{2}} u(x) \), the derivative \( f'(x) \) will thus be \( \frac{1}{\sqrt{2}} \cdot u'(x) \). Substitute \( u'(x) \) from Step 2 to get the derivative: \[ f'(x) = \frac{1}{\sqrt{2}} \cdot \frac{15x^2 + 5}{2\sqrt{5x^3 + 5x}} \].

Step 4: Simplify the Expression

Combine and simplify the constants and expressions if possible. The final derivative becomes \[ f'(x) = \frac{15x^2 + 5}{2\sqrt{2} \cdot \sqrt{5x^3 + 5x}} = \frac{15x^2 + 5}{2\sqrt{10x^3 + 10x}}. \]

Key Concepts

Chain Rule

The chain rule is essential for differentiating complex functions. It's used when you have a function inside another function. In other words, when you want to differentiate the composition of two functions.
To apply the chain rule effectively, you must identify both the outer function and the inner function. The rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function multiplied by the derivative of the inner function itself.

• If you have a function of the form \( g(f(x)) \), then the derivative is \( g'(f(x)) \cdot f'(x) \).
• Always take the derivative of the outer function, leaving the inner function as it is, and then multiply by the derivative of the inner function.

In the original exercise, when differentiating \( u(x) = \sqrt{5x^3 + 5x} \), the inner function is \( 5x^3 + 5x \), and its derivative is \( 15x^2 + 5 \). You then multiply by the derivative of the inner part, which is captured using the chain rule.

Product Rule

The product rule is another differentiation technique useful in situations where two functions are multiplied together. In calculus, this rule simplifies finding the derivative of a product of two functions.
The product rule states that if you have two functions, \( u(x) \) and \( v(x) \), their product \( u(x) \cdot v(x) \) has a derivative:

• \( (uv)' = u'v + uv' \)

This means you take the derivative of the first function multiplied by the second function as it is, plus the first function as it is multiplied by the derivative of the second function. In the exercise, while you might expect needing the product rule, since the main operation is involving a constant and a function, only the constant multiple rule was necessary. However, understanding the product rule can clarify when to apply it in more complex cases.

Differentiation Techniques

Differentiation techniques are essential for handling various types of functions you face in calculus. These techniques include the chain rule, product rule, quotient rule, and others, each catering to specific functional patterns.
Here's a quick overview:

• Chain Rule: Useful for nested functions, where you differentiate the outer function and multiply by the derivative of the inner function.
• Product Rule: Applicable when differentiating a product of two functions. Remember the formula: \((uv)' = u'v + uv'\).
• Quotient Rule: Essential when differentiating a division of two functions. The formula is \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \).
• Constant Multiple Rule: If a function is multiplied by a constant, differentiate as usual and then multiply the result by that constant.

Applying the right technique makes calculating derivatives more efficient and less error-prone. In the exercise, different rules were combined to achieve the final result, showcasing the importance of mastering these techniques for solving calculus problems effectively.