Problem 4 - Calculus for Biology and Medicine

Question

Differentiate the functions with respect to the independent variable. \(f(x)=\left(5 x^{2}-3 x\right)^{2}\)

Step-by-Step Solution

Answer

The derivative is \( f'(x) = 100x^3 - 90x^2 + 18x \).

Step 1: Identify the Rule to Apply

The function is a composite function in the form of
7. We can use the chain rule for differentiation in order to find the derivative of this type of function.

Step 2: Apply the Chain Rule

First, identify the outer function and inner function. Here, the outer function is
7. According to the chain rule, if we have functions \( u(x) \) and \( v(x) \), where \( u = g(v) \), then the derivative is \( f'(x) = (2 (5x^2 - 3x)) \cdot \frac{d}{dx}(5x^2 - 3x) \).

Step 3: Differentiate the Inner Function

Compute the derivative of the inner function, which is \( u(x) = 5x^2 - 3x \). Use basic differentiation rules: \( \frac{d}{dx}(5x^2) = 10x \) and \( \frac{d}{dx}(-3x) = -3 \). Therefore, \( \frac{d}{dx}(5x^2 - 3x) = 10x - 3 \).

Step 4: Apply the Derivatives

Substitute the derivative of the inner function obtained in Step 3 into the expression derived in Step 2: \( f'(x) = 2(5x^2 - 3x)(10x - 3) \).

Step 5: Simplify the Expression

Distribute the terms to find the expanded derivative: \( f'(x) = 2 \cdot (5x^2 \cdot 10x - 5x^2 \cdot 3 - 3x \cdot 10x + 3x \cdot 3) \). Simplify the expression to get the final derivative: \( f'(x) = 100x^3 - 30x^2 - 60x^2 + 18x \), resulting in \( f'(x) = 100x^3 - 90x^2 + 18x \).

Key Concepts

Composite Function

When dealing with mathematics, especially in calculus, one may often encounter what are called composite functions. Composite functions are essentially functions within functions. Imagine you have two functions: \( f(x) \) and \( g(x) \). If you nest one inside the other, like \( f(g(x)) \), you've got yourself a composite function.

They play a crucial role in calculus because many real-world situations can be modeled by composite functions, where one variable relies on another variable's outcome.

• A solid grasp of composite functions is necessary for tackling various calculus problems, particularly when you need to differentiate them using other rules like the chain rule.
• Understanding how to manage these nested functions is essential for predicting the behavior of the composite in different scenarios.

Identifying a composite function is the first step in many differentiation problems, as we saw in the example where the function \( f(x)=(5x^2 - 3x)^2 \) was clearly a composite.

Derivative

The derivative of a function describes how the function's output changes as the input changes. In simpler terms, it tells you the rate at which something is changing. Let's say we have a function \( y = f(x) \), the derivative, represented as \( f'(x) \) or \( \frac{dy}{dx} \), gives us insight into the rate of change of \( y \) with respect to \( x \).

• Derivatives are foundational because they allow us to determine things like slopes of curves and rates of change in various disciplines, be it physics, economics, or engineering.
• Understanding derivatives involves recognizing how to apply rules of differentiation to find the change rate of composite functions.

In the given example, differentiating \( f(x) = (5x^2 - 3x)^2 \) required us to find the rate of change by leveraging the chain rule. We first found the derivative of the inside function followed by the outside.

Differentiation Rules

To differentiate functions comfortably, it's important to understand and apply various differentiation rules. These rules provide the framework for solving complex differentiation problems with ease. Some of the key rules include:

• Power Rule: Used when differentiating terms like \( x^n \), where the derivative is \( nx^{n-1} \).
• Chain Rule: Utilized specifically for differentiating composite functions. If \( y = g(h(x)) \), then the derivative is \( y' = g'(h(x)) \, h'(x) \).
• Product and Quotient Rules: Important when dealing with products and ratios of functions. The product rule states \( (uv)' = u'v + uv' \) while the quotient is \( (\frac{u}{v})' = \frac{u'v - uv'}{v^2} \).

Each of these rules is indispensable when it comes to differentiating complex functions like the one in our example, where the chain rule was highlighted to successively differentiate composite structures.