Problem 65 - Calculus and its applications

Question

For \(y=x^{k},\) find \(d^{5} y / d x^{5}\)

Step-by-Step Solution

Answer

The fifth derivative is \(\frac{d^5 y}{dx^5} = k(k-1)(k-2)(k-3)(k-4)x^{k-5}\).

Step 1: Differentiate once

Start by differentiating the function \(y = x^k\). The derivative is found using the power rule, which states that \(d/dx \, [x^n] = n x^{n-1}\). Therefore, \(dy/dx = k x^{k-1}\).

Step 2: Differentiate a second time

Differentiate the first derivative again using the power rule. This gives:\[\frac{d^2 y}{dx^2} = \frac{d}{dx} (k x^{k-1}) = k(k-1) x^{k-2}\]

Step 3: Differentiate a third time

Take the derivative of the second derivative:\[\frac{d^3 y}{dx^3} = \frac{d}{dx} (k(k-1) x^{k-2}) = k(k-1)(k-2) x^{k-3}\]

Step 4: Differentiate a fourth time

Now differentiate the third derivative:\[\frac{d^4 y}{dx^4} = \frac{d}{dx} (k(k-1)(k-2) x^{k-3}) = k(k-1)(k-2)(k-3) x^{k-4}\]

Step 5: Differentiate a fifth time

Differentiate the fourth derivative to reach the fifth derivative:\[\frac{d^5 y}{dx^5} = \frac{d}{dx} (k(k-1)(k-2)(k-3) x^{k-4}) = k(k-1)(k-2)(k-3)(k-4) x^{k-5}\]

Key Concepts

Power Rule

The power rule is a quick and efficient way to differentiate expressions of the form \(x^n\). It is a fundamental technique in calculus that simplifies the process of finding derivatives. The rule states that if you have a function \(y=x^n\), then its derivative with respect to \(x\) is found by multiplying the exponent \(n\) by the base \(x\) raised to the power of \(n-1\). This can be expressed as

• \( \frac{d}{dx}[x^n] = n \cdot x^{n-1} \)

To see why this works, consider the case where \(n=3\). Applying the power rule would mean the derivative of \(x^3\) becomes \(3x^{2}\). This simple formula makes it incredibly easy to handle polynomial functions and find higher-order derivatives.
By repeatedly applying the power rule, we can find the derivatives needed at each step for increasing levels of differentiation, as seen in the exercise.

Differentiation

Differentiation is one of the core operations in calculus, and it allows you to find the rate of change of a function. When you differentiate a function, you are essentially finding how the function's output value changes with respect to changes in its input value. In mathematical terms, differentiation gives you the derivative.
In practical terms, think of a car's speed at a specific moment; that speed is the rate of change of the car's position over time.

• For example, the derivative of \(y = x^k\) is \(dy/dx = kx^{k-1}\).
• This provides an understanding of how fast \(y\) changes for a small change in \(x\).

The process of differentiation revolves around applying rules like the power rule and involves operations over multiple steps, especially for higher-order derivatives. As seen in the given exercise, when calculating derivatives up to the fifth order, each step involves applying the power rule systematically to find new expressions representing the rate of change at each level.

Calculus Problem Solving

Solving calculus problems often involves breaking down complex operations into simpler, more manageable steps. The example provided is a great illustration of this process. When tasked with finding higher-order derivatives, the key strategy is to apply known rules, such as the power rule, in sequence.

• Start by considering the initial function, \(y = x^k\), and apply the power rule to find the first derivative.
• Then, treat the first derivative as a new function and apply the power rule again to find the second derivative.
• Continue this process of reapplying differentiation to each subsequent derivative, step by step, until the desired order is achieved.

This stepwise approach is essential because it breaks down what could be a daunting task into smaller, simpler parts.
By focusing on each derivative individually, you'll avoid mistakes and ensure a correct, accurate result. This methodical way of thinking is a crucial part of calculus problem solving and is widely applicable to various types of calculus challenges you might encounter.