Problem 14 - Single Variable Calculus

Question

Find the derivative of the function. \(y=-\frac{1}{3}\left(x^{3}+2 x^{2}+x-1\right)\)

Step-by-Step Solution

Answer

\(y' = -x^2 - \frac{4}{3}x - \frac{1}{3}\)

Step 1: Identify and apply the differentiation rules

We will be using the power rule, which states that the derivative of x^n with respect to x is nx^(n-1), and the constant multiple rule, which says that the constant multiples can be taken out of differentiation. The given function is: \(y= -\frac{1}{3}(x^3 + 2x^2 + x - 1)\) First, we differentiate each term inside the parentheses with respect to x.

Step 2: Differentiate the first term (x^3)

Applying the power rule to the first term x^3, we get: \(\frac{d}{dx}x^3 = 3x^2\)

Step 3: Differentiate the second term (2x^2)

Applying the power rule to the second term 2x^2, we get: \(\frac{d}{dx}2x^2 = 4x\)

Step 4: Differentiate the third term (x)

Applying the power rule to the third term x, we get: \(\frac{d}{dx}x = 1\)

Step 5: Differentiate the fourth term (-1)

Since the fourth term is a constant, its derivative is zero: \(\frac{d}{dx}(-1) = 0\)

Step 6: Combine the derivatives and apply the constant multiple rule

Now, we combine the derivatives we found in the above steps inside the parentheses: \(\frac{d}{dx}(x^3 + 2x^2 + x - 1) = 3x^2 + 4x + 1 - 0 = 3x^2 + 4x + 1\) Next, we apply the constant multiple rule and multiply our result by -1/3: \(y' = -\frac{1}{3}(3x^2 + 4x + 1)\)

Step 7: Simplify the final result (optional)

Our final derivative is: \(y' = -x^2 - \frac{4}{3}x - \frac{1}{3}\) We have successfully found the derivative of the given function.

Key Concepts

Power Rule in Calculus

The power rule is a fundamental principle in calculus that simplifies the differentiation of polynomial expressions. It's particularly useful when you need to find the derivative of a function in the form of \(x^n\). To apply the power rule, multiply the power \(n\) by the coefficient, and then subtract one from the power. For instance, if you have \(x^3\), the derivative is calculated as \(3x^{3-1} = 3x^2\). This method not only applies to terms where the exponent is an integer but is also applicable to fractional or negative exponents, expanding its usefulness.This rule is paramount for students because many functions in basic calculus courses are presented in polynomial form, making the power rule indispensable for solving such problems efficiently. Remember:

• Identify the exponent \(n\).
• Multiply by \(n\), then reduce the exponent by one.

With practice, applying the power rule becomes almost second nature.

Understanding the Constant Multiple Rule

The constant multiple rule is another key rule in differentiation that pairs well with the power rule. It states that if you have a constant multiplied by a function, you can "carry out" the constant and focus on differentiating the remaining part. For the function \(-\frac{1}{3}(x^3 + 2x^2 + x - 1)\), the constant \(-\frac{1}{3}\) can be factored out, allowing you to differentiate \((x^3 + 2x^2 + x - 1)\) independently.Here's how it works:

• Identify the constant multiplier outside the derivative.
• Differentiate the rest of the function as usual.
• Multiply the differentiated result back by the constant.

This rule saves time and simplifies work, especially when dealing with scalar multiples, helping you focus on more complex aspects of a calculus problem.

A Deep Dive into Differentiation

Differentiation is a core concept in calculus that involves finding the derivative, or rate of change, of a function. It indicates how a function's output value changes as its input changes. In our exercise, differentiation helps understand how the function \(y= -\frac{1}{3}(x^3 + 2x^2 + x - 1)\) behaves as \(x\) varies. Differentiation is applied through rules like the power rule and the constant multiple rule, providing tools to derive more complex functions easily. It is crucial in:

• Calculating slopes of curves in graphical analysis.
• Maximizing or minimizing values in optimization problems.
• Understanding instantaneous rates of change in physical contexts.

For single-variable functions, differentiation usually entails dealing with polynomials and utilizing basic rules to compute derivatives effectively.

Single Variable Calculus and Its Importance

Single variable calculus focuses on functions of one independent variable and their derivatives and integrals. The exercise on deriving \(y= -\frac{1}{3}(x^3 + 2x^2 + x - 1)\) falls into this category, as we compute the derivative concerning the single variable \(x\).Key aspects include:

• Mastery of basic differentiation rules such as the power and constant multiple rules.
• Analysis of polynomial functions and their behavior.
• Application of differentiation to solve real-world problems involving rates of change.

In single-variable calculus, students learn to handle problems with increasing complexity using fundamental concepts they initially acquire. Understanding these basics creates a solid foundation for exploring more advanced calculus topics, including multivariable calculus.