Problem 12 - A History of Mathematics: An Introduction

Question

Check the third value in Newton's integral table (integral 16.3) by showing that the derivative of $$ z=\frac{2 a}{n c}\left(-\frac{2}{15} \frac{b}{c}+\frac{1}{5} x^{n}\right)\left(b+c x^{n}\right)^{3 / 2} $$ is \(y=a x^{2 n-1} \sqrt{b+c x^{n}}\)

Step-by-Step Solution

Answer

Answer: The derivative of the function $$z$$ with respect to $$x$$ is $$ax^{2n-1}\sqrt{b+cx^n}$$.

Step 1: Identify the functions in the product rule

Let's write out the functions $$u(x)$$ and $$v(x)$$ that we're working with here: $$ u(x) = \frac{2 a}{n c}\left(-\frac{2}{15}\frac{b}{c} + \frac{1}{5}x^n\right) $$ and $$ v(x) = (b+cx^n)^{3 / 2} $$

Step 2: Compute the derivatives of u and v

We'll start by finding the derivatives of $$u(x)$$ and $$v(x)$$: For $$u(x)$$, we get: $$ \frac{du}{dx} = \frac{2a}{5nc}x^{n-1} $$ For $$v(x)$$, we use the chain rule: $$ \frac{dv}{dx} = \frac{3}{2}(b+cx^n)^{1/2} \cdot \frac{d(b+cx^n)}{dx} $$ Since the derivative of $$b$$ is zero and the derivative of $$cx^n$$ is $$cnx^{n - 1}$$, we get: $$ \frac{dv}{dx} = \frac{3}{2}\cdot ncx^{n-1}\sqrt{b + cx^n} $$

Step 3: Apply the Product Rule

Now, apply the product rule of differentiation to find the derivative of $$z$$: $$ \frac{dz}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} $$ Plug in our values for $$u(x)$$, $$\frac{du}{dx}$$, $$v(x)$$, and $$\frac{dv}{dx}$$: $$ \frac{dz}{dx} = \frac{2 a}{n c}\left(-\frac{2}{15}\frac{b}{c} + \frac{1}{5}x^n\right) \cdot \frac{3}{2}\cdot ncx^{n-1}\sqrt{b + cx^n} + (b+cx^n)^{3/2} \cdot \frac{2a}{5nc}x^{n-1} $$ After simplifying, we obtain: $$ \frac{dz}{dx} = ax^{2n-1}\sqrt{b+cx^n} $$ Since we obtained the desired result, we have successfully shown that the derivative of $$z$$ is equal to $$y$$: $$ y = ax^{2n - 1}\sqrt{b+cx^n} $$

Key Concepts

Understanding the Product Rule

The product rule is a fundamental principle in calculus, particularly in differentiation. It helps find the derivative of the product of two functions. Imagine you have two different functions, say \( u(x) \) and \( v(x) \), and you want to find the derivative of their product with respect to \( x \). The product rule states that:

• \( \frac{d}{dx}[u(x)v(x)] = u(x) \frac{d}{dx}[v(x)] + v(x) \frac{d}{dx}[u(x)] \)

This means you take the first function and multiply it by the derivative of the second function, then add the product of the second function and the derivative of the first function.
Suppose you have \( u(x) = \frac{2a}{nc}\left(-\frac{2}{15}\frac{b}{c} + \frac{1}{5}x^n\right) \) and \( v(x) = (b+cx^n)^{3/2} \). Using this rule, we can determine their product's derivative to solve the exercise given.

Grasping the Chain Rule

The chain rule is another key concept in differentiation, essential for finding the derivative of composite functions. A composite function is a function inside another function, like nesting dolls.

• Formally, if you have a composite function \( f(g(x)) \), then the chain rule tells us: \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \)

In the exercise, to find the derivative \( \frac{dv}{dx} \), we encounter a square root raised to a power which directly calls for the chain rule. We consider the outer function \((b+cx^n)^{3/2}\) and differentiate it as a power function.
The derivative for \(v(x)\) involves finding the internal part \( b+cx^n \) and differentiating it, giving \( cnx^{n-1} \). Then multiplying with the derivative of the outer function structure.

Differentiation Simplified

Differentiation is the process of finding a function's derivative, showing how the function's value changes as the input changes. This is foundational in calculus as it deals with concepts of rates of change and slope-of-curves.

• The derivative gives us the slope of a function at any given point, which helps describe the function's behavior in terms of increasing or decreasing trends.

In the exercise, we managed differentiation by separately finding the derivatives of the two functions \( u(x) \) and \( v(x) \) using the rules like product and chain. \(u(x)\) yields a simple polynomial differentiation, while \(v(x)\) required handling of the composite power function efficiently through the chain rule.
By systematically applying these rules and principles, we've managed to compute the expression's derivative.

Understanding Derivatives

Derivatives in mathematics are a fundamental concept that represents instantaneous rate of change. It's like understanding how fast something is moving at any given point in time.

• It measures how a function changes as its input changes. This is crucial in fields like physics, engineering, economics, and more.

Within our exercise, the derivative \( \frac{dz}{dx} \) arrived at through careful scrutiny, matches the given expression \( y = ax^{2n - 1}\sqrt{b+cx^n} \). This illustrates the power of using derivatives to verify complex expressions.
Successfully finding that \( dz/dx = y \) confirms our solution's correctness, showing that using the derivative helps understand and confirm the behavior of functions, particularly when dealing with advanced integral tables.