Question

In Exercises, find the second derivative of the function. $$ y=\left(x^{3}-2 x\right)^{4} $$

Short answer

The steps for solving this involve the power rule, the chain rule, and the product rule of differentiation. After applying these rules to find the first and second derivative, they should be simplified to their simplest form.

Step-by-step solution

Find the first derivative

Use the power rule of differentiation, which says that the derivative of \(x^n\) is \(nx^{n-1}\), and the chain rule, which says that the derivative of \(f(g(x))\) is \(f'(g(x))g'(x)\). In this case, you get \(y' = 4(x^3 - 2x)^3(3x^2 - 2)\).

Simplify the first derivative

Simplify the derivative by multiplying inside the brackets \(y' = 12(x^3 - 2x)^3x^2 - 8(x^3 - 2x)^3\).

Find the second derivative

To find the second derivative, take the derivative of the first derivative. Apply the product and chain rule again, this step will require careful application of the product, chain and power rules.

Simplify the second derivative

Simplify the derivation result, try to factor out similar terms and write the second derivative in the simplest form.

Key concepts

Chain Rule

The Chain Rule is a fundamental technique in calculus for handling compositions of functions. It tells us how to differentiate a function that is composed of two or more functions.
Simply put, if you have a function like \((f \circ g)(x) = f(g(x))\), then the Chain Rule states that its derivative is given by:

• Take the derivative of the outer function \(f\) with respect to \(g(x)\), noted as \(f'(g(x))\).
• Multiply this by the derivative of the inner function \(g(x)\) with respect to \(x\), noted as \(g'(x)\).

So, the formula will look like this: \[ (f \circ g)'(x) = f'(g(x)) \cdot g'(x) \]In the context of differentiating \(y = (x^3 - 2x)^4\), we treated \((x^3 - 2x)\) as the inner function and \((\cdot)^4\) as the outer function. The application of the Chain Rule made it possible to tackle the complexity of the derivative with ease.

Power Rule

The Power Rule is one of the most straightforward and commonly used rules in calculus for differentiation.
It states that if you have a function of the form \(x^n\), then its derivative is \(nx^{n-1}\).Consider a simple example: for the function \(x^4\), applying the Power Rule would give us a derivative of \(4x^3\).In our exercise, when you encounter a component like \((x^3 - 2x)^4\), the Power Rule helps us to handle it by taking the exponent down and reducing it by one. This is captured in the expression \(4(x^3 - 2x)^3\), which emerges from differentiating the outer function in combination with the Chain Rule.If you become comfortable with the Power Rule, it significantly simplifies the process of differentiation, especially in complex problems.

Product Rule

The Product Rule is central when differentiating the product of two functions. If you have two functions \(u(x)\) and \(v(x)\), then the Product Rule states:

• The derivative of their product \(u(x) \cdot v(x)\) is \(u'(x) \cdot v(x) + u(x) \cdot v'(x)\).

In simple terms, differentiate the first function and multiply by the second, then add the first function times the derivative of the second.For our exercise, we use the Product Rule to differentiate terms like \(12(x^3 - 2x)^3x^2 - 8(x^3 - 2x)^3\). Each term involves functions multiplied together, which is the perfect scenario for employing the Product Rule.Understanding how to properly execute the Product Rule allows for solving more intricate problems where multiplication of differentiable functions occurs.

Differentiation

Differentiation is a critical operation in calculus that deals with finding a derivative, which signifies the rate of change of a function. It allows us to understand how the function behaves and how its values change over its domain. Derivation depends heavily on rules such as the Chain Rule, Power Rule, and Product Rule to break down complicated expressions.
For example, when finding the second derivative of a function like the one in our exercise, you'll need to carefully apply multiple differentiation rules. First, we find the first derivative and then differentiate it again to obtain the second derivative. Differentiation is a powerful tool, enabling you to determine tangents to curves, rates of change, and much more within various mathematical contexts.

Simplification

Simplification is about making complex expressions easier to manage and understand. After obtaining a derivative, the expression can be messy and daunting to interpret.

• Combine like terms to reduce the expression to simpler forms.
• Factor out common factors which can help to reveal underlying patterns.

For instance, from our given exercise, simplifying the derivatives helped us express the result succinctly and understandably. Simplifying isn't just about aesthetics; it's crucial for verifying the accuracy of the solution and making further calculations or interpretations more accessible. Being proficient at simplifying ensures that your mathematical expressions are always as clear and as concise as possible.