Problem 21 - Calculus - AP Edition

Question

$$\text { Find } d y /\left.d x\right|_{x=1}$$ $$y=1+x+x^{2}+x^{3}+x^{4}+x^{5}$$

Step-by-Step Solution

Answer

The derivative at \(x = 1\) is 15.

Step 1: Write down the given function

The given function is: \[ y = 1 + x + x^2 + x^3 + x^4 + x^5 \]

Step 2: Differentiate the function

To find the derivative \( \frac{dy}{dx} \), differentiate each term of the function with respect to \(x\). - The derivative of 1 is 0. - The derivative of \(x\) is 1. - The derivative of \(x^2\) is \(2x\). - The derivative of \(x^3\) is \(3x^2\). - The derivative of \(x^4\) is \(4x^3\). - The derivative of \(x^5\) is \(5x^4\). Thus, \[ \frac{dy}{dx} = 1 + 2x + 3x^2 + 4x^3 + 5x^4 \]

Step 3: Evaluate the derivative at specified point

To find \( \left. \frac{dy}{dx} \right|_{x=1} \), substitute \(x = 1\) into the derivative:\[ \frac{dy}{dx} = 1 + 2(1) + 3(1)^2 + 4(1)^3 + 5(1)^4 \] Simplify this expression.

Step 4: Simplify the expression

Calculate each term:- \(2(1) = 2\)- \(3(1)^2 = 3\)- \(4(1)^3 = 4\)- \(5(1)^4 = 5\)Substitute back into the expression:\[ 1 + 2 + 3 + 4 + 5 \] This simplifies to 15.

Key Concepts

Polynomials

A polynomial is a mathematical expression involving a sum of powers in one or more variables multiplied by coefficients. In simpler terms, it's an equation composed of variables raised to integer powers and those powers are usually expressed starting from zero up to a certain positive integer. Polynomials can be neatly organized using some basic rules.

Let's look at the polynomial from the original exercise:

• The function, given as a polynomial, is: \( y = 1 + x + x^2 + x^3 + x^4 + x^5 \).
• Here, \(1, x, x^2, x^3, x^4, x^5\) are terms of the polynomial.
• The numbers multiplying the powers of \(x\) are called coefficients.
• This polynomial is of degree 5 because the highest power of \(x\) is 5.

Understanding polynomials is crucial because they appear in various areas of algebra and play a significant role in calculus, particularly when finding derivatives and integrations.

Differentiation

Differentiation is the process of finding the derivative of a function. A derivative represents the rate of change of one quantity with respect to another. It is a fundamental tool in calculus that helps understand how a function changes at any given point.

To differentiate a polynomial, you apply the power rule, which states that if you have a function \(x^n\), its derivative is \(nx^{n-1}\). Let's apply this to the example function:

• The derivative of 1 is 0 because constants have no rate of change.
• The derivative of \(x\) is 1.
• The derivative of \(x^2\) is \(2x\).
• The derivative of \(x^3\) is \(3x^2\).
• The derivative of \(x^4\) is \(4x^3\).
• The derivative of \(x^5\) is \(5x^4\).

Putting it all together, the derivative of the function \(y = 1 + x + x^2 + x^3 + x^4 + x^5\) is \(\frac{dy}{dx} = 1 + 2x + 3x^2 + 4x^3 + 5x^4\). This tells us how the values of \(y\) change as \(x\) changes, with each term contributing differently to the overall change.

Calculus

Calculus is the branch of mathematics that deals with the study of change. It is divided into differential calculus and integral calculus, and it provides various tools for analyzing the behavior of functions in terms of their rates of change and total accumulation.

In the exercise, the aspect of calculus we focus on is differentiation, which is a key concept in differential calculus. The overall goal was to evaluate the derivative at a specific point, \(x = 1\). This involves substituting \(x = 1\) into the differentiated function. Here's how that looks:

• Calculate \(1 + 2(1) + 3(1)^2 + 4(1)^3 + 5(1)^4\).
• This simplifies, with each term reducing to its coefficient: \(1 + 2 + 3 + 4 + 5\).
• The sum of these terms is 15, which is the rate of change of the function at \(x=1\).

This shows how calculus can be used to determine the specific behavior of a function. At \(x=1\), the function increases by 15 per unit change in \(x\), demonstrating how crucial calculus is in both theoretical and applied mathematics.