Problem 29 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

\(y=\frac{\left(x^{2}+e^{2 x}\right)^{3} e^{-2 x}}{\left(1+x-x^{2}\right)^{2 / 3}}\)

Step-by-Step Solution

Answer

Use logarithmic differentiation and simplify: \[ \frac{dy}{dx} = \frac{(x^2 + e^{2x})^3 e^{-2x}}{(1 + x - x^2)^{2/3}} \bigg[ \frac{6x + 6e^{2x}}{x^2 + e^{2x}} - 2 - \frac{2(1-2x)}{3(1+x-x^2)} \bigg] \]

Step 1: Understand the Expression

The given expression to differentiate is: \[ y = \frac{\big(x^2 + e^{2x}\big)^3 e^{-2x}}{\big(1 + x - x^2\big)^{2/3}} \]

Step 2: Use Logarithmic Differentiation

To simplify differentiation, use logarithmic differentiation. Let \( y = f(x) \) Take the natural logarithm of both sides: \[ \text{ln}(y) = \text{ln}\bigg(\frac{\big(x^2 + e^{2x}\big)^3 e^{-2x}}{\big(1 + x - x^2\big)^{2/3}}\bigg) \]

Step 3: Simplify Using Logarithmic Properties

Using logarithm properties, break down the expression: \[\text{ln}(y) = \text{ln}\big((x^2 + e^{2x})^3\big) + \text{ln}(e^{-2x}) - \text{ln}\big((1 + x - x^2)^{2/3}\big)\] Simplify further: \[ \text{ln}(y) = 3\text{ln}(x^2 + e^{2x}) - 2x - \frac{2}{3}\text{ln}(1 + x - x^2) \]

Step 4: Differentiate Both Sides

Differentiate both sides with respect to \( x \): \[ \frac{d}{dx}[\text{ln}(y)] = \frac{d}{dx}\bigg[3\text{ln}(x^2 + e^{2x}) - 2x - \frac{2}{3}\text{ln}(1 + x - x^2)\bigg] \] This gives us: \[ \frac{1}{y} \frac{dy}{dx} = 3 \frac{1}{x^2 + e^{2x}} \bigg(2x + 2e^{2x}\bigg) - 2 - \frac{2}{3} \frac{1}{1 + x - x^2} (1 - 2x) \]

Step 5: Simplify the Derivative

Simplify the result from Step 4: \[ \frac{dy}{dx} = y \bigg[ \frac{3(2x + 2e^{2x})}{x^2 + e^{2x}} - 2 - \frac{2 (1 - 2x)}{3 (1 + x - x^2)} \bigg] \]

Step 6: Substitute Back for y

Recall that \( y = \frac{(x^2 + e^{2x})^3 e^{-2x}}{(1 + x - x^2)^{2/3}} \).Substitute this back into the expression for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{(x^2 + e^{2x})^3 e^{-2x}}{(1 + x - x^2)^{2/3}} \bigg[ \frac{6x + 6e^{2x}}{x^2 + e^{2x}} - 2 - \frac{2 (1 - 2x)}{3 (1 + x - x^2)} \bigg] \]

Key Concepts

Natural Logarithm

Understanding the natural logarithm (\text{ln}) is crucial for simplifying complex functions. The natural logarithm is the logarithm to the base of the mathematical constant \(e\) (approximately equal to 2.71828). In logarithmic differentiation, we take the natural logarithm of both sides of an equation to make differentiation easier. For example, when we have a function \(y = \frac{\big(x^2 + e^{2x}\big)^3 e^{-2x}}{\big(1 + x - x^2\big)^{2/3}}\), we apply \(\text{ln}\) to both sides: \(\text{ln}(y) = \text{ln}\big(\frac{\big(x^2 + e^{2x}\big)^3 e^{-2x}}{\big(1 + x - x^2\big)^{2/3}}\big)\).

Using properties of logarithms such as \(\text{ln}(ab) = \text{ln}(a) + \text{ln}(b)\) and \(\text{ln}(a^b) = b\text{ln}(a)\), we can break down and simplify complex expressions more easily. This aids in the next step, differentiation.

Differentiation

Differentiation is a fundamental concept in calculus where we find the derivative of a function, representing the rate of change. Logarithmic differentiation involves differentiating implicit functions and fractions more easily by first converting the problem using logarithms.

After applying the natural logarithm to our function and simplifying, we differentiate both sides with respect to \(x\). For example, if \(\text{ln}(y) = 3\text{ln}(x^2 + e^{2x}) - 2x - \frac{2}{3}\text{ln}(1 + x - x^2)\), we differentiate to get: \(\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}[3\text{ln}(x^2 + e^{2x})] - \frac{d}{dx}[2x] - \frac{d}{dx}[\frac{2}{3}\text{ln}(1 + x - x^2)]\).

This simplifies to \(\frac{dy}{dx} = y \big[ \frac{6x + 6e^{2x}}{x^2 + e^{2x}} - 2 - \frac{2 (1 - 2x)}{3 (1 + x - x^2)} \big]\).

Calculus

Calculus is a branch of mathematics focusing on limits, functions, derivatives, integrals, and infinite series. It has two main branches: differential calculus and integral calculus. Differential calculus is concerned with the concept of a derivative, while integral calculus deals with integration.

In this particular problem, we used differential calculus techniques, specifically logarithmic differentiation, to find the derivative of a complex function involving exponential, polynomial, and fractional terms. The steps were:

• Taking the natural logarithm of both sides of the equation to leverage logarithmic properties.
• Simplifying the logarithmic expression.
• Applying the rules of differentiation to each term.
• Substituting back the original function to express the final derivative.

This method is one of the many powerful tools in calculus used to simplify and solve problems that might otherwise be very difficult to handle.