Question

Finding a Pattern Develop a general rule for \([x f(x)]^{(n)},\) where \(f\) is a differentiable function of \(x .\)

Short answer

The nth derivative of \(x f(x)\) appears to be \((n f^{(n-1)}(x)) + x f^{(n)}(x)\). This pattern applies from the second derivative onwards.

Step-by-step solution

Understanding the problem

We are asked to find a pattern, or rule, for the nth derivative of the product of \(x\) and a differentiable function \(f(x)\). The first few derivatives must be computed for \(x f(x)\) to try to identify any potential patterns.

Finding the first derivative

We can start by finding the first derivative using the product rule, which gives \(f(x) + x f'(x)\).

Finding the second derivative

Next, we find the second derivative of \(xf(x)\), which is \((f(x) + x f'(x))' = f'(x) + f'(x) + x f''(x) = 2f'(x) + x f''(x)\).

Finding the third derivative

Let's try finding the third derivative, which is \((2f'(x) + x f''(x))' = 2 f''(x) + f''(x) + x f'''(x) = 3 f''(x) + x f'''(x)\).

Identifying the pattern

By observing, we can see a pattern established in this sequence of derivatives - it seems that the nth derivative of \(x f(x)\) is \((n f^{(n-1)}(x)) + xf^{(n)}(x)\). This pattern applies from the second derivative onwards. The exact form of the pattern will vary depending on the specific function \(f(x)\).

Key concepts

Product Rule

When you are dealing with differentiation, especially in the context of finding derivatives of products of two functions, the **Product Rule** is an essential tool. The Product Rule states that if you have two differentiable functions, say \(u(x)\) and \(v(x)\), the derivative of their product is given by:

• \((uv)' = u'v + uv'\)

This rule helps us break down the differentiation process into simpler components when dealing with products of functions. Remember:

• \(u'(x)\) represents the derivative of \(u(x)\)
• \(v'(x)\) represents the derivative of \(v(x)\)

In the exercise, since we have the product \(x f(x)\), we applied this rule to get the first derivative, \(f(x) + x f'(x)\). Splitting this operation into two smaller parts makes it less complex and manageable.

nth Derivative

Moving beyond the basic derivative, the **nth Derivative** is a notation used to express the derivative of a function taken \(n\) times. Derivatives can be taken further beyond the first derivative, which helps to gain additional insights about the behavior of functions. As you find successive derivatives — second, third, and beyond — each provides information such as acceleration and changes in the rate of growth or decay.

In our original problem, we identified a pattern for the derivative of \(x f(x)\) as the derivation progressed beyond a few iterations.

• For the nth derivative of \(x f(x)\), the pattern that emerges is \((n f^{(n-1)}(x)) + xf^{(n)}(x)\).

This pattern can help solve complex differentiation tasks quickly once identified, as it allows for predictability in the equation's progression.

Differentiable Function

A **Differentiable Function** is a function that has a derivative at every point in its domain. In simpler terms, the graph of a differentiable function doesn’t have breaks, cusps, or corners — it’s smooth. Differentiability lets us predict the behavior of the function based on its derivative.

• If \(f(x)\) is differentiable, it ensures that its derivatives \(f'(x), f''(x),\) and so on continue to exist.

In calculus, exploring the differentiability of a function is crucial. It shows us how to understand changes in function values, velocity, and even areas under curves. Importantly, in our task, the condition that \(f(x)\) is differentiable was necessary to find multiple derivatives of the product \(x f(x)\), ensuring the function behaves predictably at each stage of differentiation.

Calculus

**Calculus** is the branch of mathematics that deals with rates of change and accumulation. It is split into differential calculus and integral calculus. Differential calculus focuses on the concept of the derivative, which measures things like how a quantity changes with respect to another. Integral calculus, on the other hand, concerns itself with accumulation of quantities, such as areas under curves.

• The tool of calculus, specifically derivatives, forms the backbone of tasks in our original problem.

With its ability to model dynamic systems and continuous change, calculus allows us to solve real-world problems in physics, engineering, economics, and beyond. Recognizing patterns in derivatives, like the one in our problem, is part of the broader application of calculus to understand and predict complex behaviors. Whether you’re calculating the derivative of a speed-time function to find acceleration, or identifying the nth derivative pattern, calculus is an essential skill.