Problem 31
Question
Find the derivative of each function. Verify some of your results by calculator. As usual, the letters \(a, b, c, \ldots\) represent constants. Derivative of a Sum. $$y=\frac{x^{2}}{2}-\frac{x^{7}}{7}$$
Step-by-Step Solution
Verified Answer
\(y' = x - x^{6}\)
1Step 1: Recognize the Function Form
The function is in the form of a sum of terms, where each term is a power of x multiplied by a constant. Each term can be differentiated independently.
2Step 2: Differentiate the First Term
Apply the power rule of differentiation to the first term, which states that the derivative of a term in the form of \(ax^n\) is \(anx^{n-1}\). For the first term \(\frac{x^{2}}{2}\), we apply the power rule: \(\frac{d}{dx} (\frac{x^{2}}{2}) = \frac{2}{2}x^{2-1} = x\).
3Step 3: Differentiate the Second Term
Apply the power rule to the second term \(-\frac{x^{7}}{7}\). The negative sign is a constant multiple and is preserved, so \(\frac{d}{dx} (-\frac{x^{7}}{7}) = -\frac{7}{7}x^{7-1} = -x^{6}\).
4Step 4: Combine the Derivatives
Combine the derivatives of the terms to obtain the derivative of the whole function: \(y' = x - x^{6}\).
Key Concepts
Power Rule of DifferentiationDerivative of a SumCalculus ProblemsDifferentiation Techniques
Power Rule of Differentiation
The power rule is a fundamental rule in calculus for finding the derivative of a function in the form of a power of x. It is expressed as \( \frac{d}{dx}[x^n] = nx^{n-1} \), where \(n\) is a real number. In practice, this means you simply multiply the exponent \(n\) by the coefficient in front of \(x\) and then subtract one from the exponent.
For example, if you are tasked to find the derivative of \(x^5\), you apply the power rule as follows: \( \frac{d}{dx}[x^5] = 5x^{5-1} = 5x^4 \). This technique streamlines the process of differentiating polynomials and is a core element of differentiation. Understanding the power rule enables students to tackle more complex calculus problems with confidence and ease.
For example, if you are tasked to find the derivative of \(x^5\), you apply the power rule as follows: \( \frac{d}{dx}[x^5] = 5x^{5-1} = 5x^4 \). This technique streamlines the process of differentiating polynomials and is a core element of differentiation. Understanding the power rule enables students to tackle more complex calculus problems with confidence and ease.
Derivative of a Sum
When you encounter a function that is composed of a sum (or difference) of terms, the derivative of the entire function can be found by differentiating each term independently and then summing the results. This approach is based on the principle that the derivative of a sum is the sum of the derivatives.
For instance, given a function \(y = u(x) + v(x)\), the derivative \(y'\) or \( \frac{dy}{dx}\) is found by differentiating \(u(x)\) and \(v(x)\) separately: \(y' = \frac{d}{dx}[u(x)] + \frac{d}{dx}[v(x)]\). For our particular example with \(y=\frac{x^{2}}{2}-\frac{x^{7}}{7}\), the differentiation is applied to each term independently, and the derivatives are then combined to get the final result.
For instance, given a function \(y = u(x) + v(x)\), the derivative \(y'\) or \( \frac{dy}{dx}\) is found by differentiating \(u(x)\) and \(v(x)\) separately: \(y' = \frac{d}{dx}[u(x)] + \frac{d}{dx}[v(x)]\). For our particular example with \(y=\frac{x^{2}}{2}-\frac{x^{7}}{7}\), the differentiation is applied to each term independently, and the derivatives are then combined to get the final result.
Calculus Problems
Solving calculus problems often involves understanding and applying a variety of rules and techniques. The process typically starts with identifying the type of function or equation you are dealing with, such as a power function, a product of functions, or a composite function. Then, appropriate differentiation techniques are employed to find the derivative.
In the context of our exercise, the problem requires recognizing the function as a sum of power functions, then applying the power rule of differentiation to each term, and finally combining the results. Throughout calculus, it's not uncommon to solve problems that involve multiple steps and concepts, reinforcing the importance of a strong foundation in differentiation techniques.
In the context of our exercise, the problem requires recognizing the function as a sum of power functions, then applying the power rule of differentiation to each term, and finally combining the results. Throughout calculus, it's not uncommon to solve problems that involve multiple steps and concepts, reinforcing the importance of a strong foundation in differentiation techniques.
Differentiation Techniques
Differentiation is a core concept in calculus, and there are several techniques that students must master. Besides the power rule, there are methods like the product rule, for differentiating the product of two functions; the quotient rule, for the division of functions; and the chain rule, for the composition of functions.
Each technique is designed to tackle specific types of functions and algebraic structures. For example, when faced with a composite function such as \(f(g(x))\), the chain rule is your go-to tool. Understanding when and how to apply these techniques is crucial for solving a wide range of calculus problems and for long-term success in the study of mathematics.
Each technique is designed to tackle specific types of functions and algebraic structures. For example, when faced with a composite function such as \(f(g(x))\), the chain rule is your go-to tool. Understanding when and how to apply these techniques is crucial for solving a wide range of calculus problems and for long-term success in the study of mathematics.
Other exercises in this chapter
Problem 30
Find the slope of the tangent to the curve \(y=1 /(x+1)\) at \(x=2\).
View solution Problem 30
Find the derivative. $$D_{x}(7-5 x)$$
View solution Problem 31
Write the differential \(d y\) for each function. $$y=\frac{x-1}{x+1}$$
View solution Problem 31
Find the derivative of each function.. $$y=\frac{x}{x+2}$$
View solution