Problem 41 - Calculus for Biology and Medicine

Question

Find \(\frac{d y}{d x}\) by applying the chain rule repeatedly. \(y=\left(\sqrt{x^{3}-3 x}+3 x\right)^{4}\)

Step-by-Step Solution

Answer

Apply the chain rule to get \( \frac{dy}{dx} = 4(u^3) \cdot [(3(x^2-1))/(2\sqrt{x^3-3x}) + 3] \).

Step 1: Recognize the Outer Function

We start by identifying the outermost function. Here, we have a power function defined as \(y = u^4\), where \(u = \sqrt{x^3 - 3x} + 3x\). This means our first differentiation should focus on the term \(u^4\).

Step 2: Differentiate the Outer Function

Using the power rule, the derivative of \(u^4\) with respect to \(u\) is given by \(4u^3\). We also need to apply the chain rule, multiplying by the derivative of \(u\), which we have represented as \(u^3\):\[ \frac{dy}{du} = 4u^3 \]

Step 3: Identify the Inner Function

The inner function, \(u\), is \(\sqrt{x^3 - 3x} + 3x\). We need to find the derivative of \(u\) with respect to \(x\). Breaking \(u\) into two parts, \(v = \sqrt{x^3 - 3x}\) and \(w = 3x\), we first examine \(v\).

Step 4: Differentiate the Nested Inner Function

The derivative of \(v = \sqrt{x^3 - 3x}\) involves more application of the chain rule. Rewrite \(v\) as \((x^3 - 3x)^{1/2}\), and differentiate using the power rule again:\[ \frac{dv}{dx} = \frac{1}{2}(x^3 - 3x)^{-1/2} \cdot (3x^2 - 3) \]This simplifies to:\[ \frac{dv}{dx} = \frac{3(x^2 - 1)}{2\sqrt{x^3 - 3x}} \]

Step 5: Differentiate the Linear Inner Function

The second part of the inner function is linear, \(w = 3x\), and differentiates simply: \[ \frac{dw}{dx} = 3 \]

Step 6: Combine the Inner Derivatives

Substitute the derivatives of \(v\) and \(w\) into the derivative of \(u\):\[ \frac{du}{dx} = \frac{dv}{dx} + \frac{dw}{dx} = \frac{3(x^2 - 1)}{2\sqrt{x^3 - 3x}} + 3 \]

Step 7: Apply the Chain Rule

Finally, using the chain rule, substitute \(\frac{du}{dx}\) back into the derivative from Step 2:\[ \frac{dy}{dx} = 4u^3 \cdot \frac{du}{dx} = 4(\sqrt{x^3 - 3x} + 3x)^3 \left( \frac{3(x^2 - 1)}{2\sqrt{x^3 - 3x}} + 3 \right) \]

Step 8: Simplify the Expression

Simplify the final expression to get the complete derivative in terms of \(x\). While simplifying, ensure each component is correctly factored or expanded as required for a neat presentation.

Key Concepts

Calculus

Calculus is a powerful branch of mathematics that deals with continuous change. It is split into differential and integral calculus.

• Differential Calculus focuses on rates of change and slopes of curves.
• Integral Calculus deals with accumulation of quantities and the areas under and between curves.

Differential calculus, which is central to this exercise, involves the process of finding derivatives. Derivatives can help us understand how a function changes at any given point. The chain rule, which is an essential tool in calculus, allows us to differentiate composite functions effectively. This is necessary when a function is made up of two or more nested functions, as demonstrated in the original exercise.

Derivatives

Derivatives represent the rate at which one quantity changes with respect to another. They are foundational to understanding and predicting how different variables interact in calculus.
To find a derivative, we apply various rules, such as:

• Power Rule: Used for differentiating functions of the form \(x^n\).
• Product Rule: Used when differentiating the product of two functions.
• Quotient Rule: Used for the fraction of two functions.
• Chain Rule: Used for composite functions, as applied in this exercise.

Understanding the Chain Rule

The chain rule is crucial when dealing with nested functions, i.e., functions within functions. By breaking down the differentiation process into parts corresponding to outer and inner functions, we can compute derivatives effectively. The initial step is to find the derivative of the outer function, treating the inner function as a single variable. After that, you calculate the derivative of the inner function itself.
All these derivative values are multiplied together to find the overall derivative of the composite function.

Functions

In mathematics, a function is a relationship or expression involving one or more variables. Functions can come in various forms and complexities.

Types of Functions in Calculus

Functions might be as straightforward as linear functions \(f(x) = mx + b\), or as complex as polynomial and composite functions, like the one in the original exercise.

• Linear Function: A straight line represented by \(y = mx + c\).
• Polynomial Function: A sum of terms, each consisting of a variable raised to a power, multiplied by a coefficient.
• Composite Function: A function applied to another function, expressed as \(f(g(x))\).

A composite function involves an `outer` function and an `inner` function. In our case, the outer function is the fourth power of the inner function, which consists of the sum of a square root and a linear term. Understanding the nature of these functions helps to apply the chain rule for derivatives efficiently, as it elucidates the relationship between the different parts of the function.