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Find the derivative of the following functions. $$g(x)=6 x^{5}-x$$

Short Answer

Expert verified
Answer: The derivative of the function g(x) = 6x^5 - x is g'(x) = 30x^4 - 1.

Step by step solution

01

Identify the power rule

The power rule is a formula used to find the derivative of a function with a power of x. The power rule states that for any function g(x) = x^n, the derivative g'(x) = nx^(n-1). In this problem, we will apply the power rule to each term in the given function.
02

Apply the power rule to the first term

Applying the power rule to the first term, we have: $$\frac{d}{dx}(6x^5) = 6\cdot5x^{(5-1)} = 30x^4$$
03

Apply the power rule to the second term

Applying the power rule to the second term, we have: $$\frac{d}{dx}(-x) = -1x^{(1-1)} = -1x^0 = -1$$
04

Combine the results

Now that we have found the derivative of each term in the function, we need to combine them to get the derivative of the entire function. Therefore, the derivative of g(x) is: $$g'(x) = 30x^4 - 1$$ This is the solution to the given exercise.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Rule
The power rule is a fundamental principle in calculus used for finding the derivative of functions with exponents. In layman's terms, the power rule gives us a shortcut so that we don't have to resort to the limit definition of the derivative every single time we want to differentiate a power function. Let's break it down: if you have a function in the form of \(f(x) = ax^n\), where \(a\) is a constant and \(n\) is a real number, the power rule tells us that the derivative of \(f\) with respect to \(x\) is \(f'(x) = anx^{n-1}\).

In the context of the given exercise, we are looking at \(g(x)=6x^5-x\). Applying the power rule, we assess each term individually. The term \(6x^5\) derives down to \(30x^4\) by multiplying the exponent 5 by the coefficient 6 and then subtracting 1 from the exponent. The term \(-x\) simplifies to \(-1\) because any variable to the power of one, when differentiated, drops the variable and retains the coefficient. Thus, understanding the power rule improves efficiency and accuracy when solving calculus problems involving derivatives of power functions.
Calculus
Calculus is an area of mathematics that deals with change and motion. It's split into two major branches: differential calculus and integral calculus. Differential calculus—which is what we're focusing on when we talk about derivatives—concerns itself with the rate at which quantities change, which is fundamental to understanding motion, force, and many other concepts. Integral calculus, on the other hand, focuses on accumulation of quantities, like areas under curves.

In solving the exercise \(g(x)=6x^5-x\), we tap into differential calculus to find the rate at which \(g(x)\) changes with respect to \(x\). Calculus is not just about mechanical computations; it helps us understand the world in motion, optimize systems, and even predict future events by analyzing rates of change. Learning calculus is akin to being given a toolkit for dissecting complex problems across various fields like physics, engineering, economics, and beyond. The beauty of the power rule is that it is a tool within this toolkit, designed to make the process of differentiation—a core operation in calculus—much more straightforward and intuitive.
Differentiation
Differentiation is the act of finding a derivative, which is essentially the rate at which one quantity changes with respect to another. When students first learn about differentiation, they often start with simpler functions and apply rules of differentiation—like the power rule—to find the derivatives. As they progress, they will encounter more complex functions and learn additional rules to tackle them effectively.

In the exercise provided, we differentiate \(g(x) = 6x^5 - x\) by applying the power rule to each term. Differentiation in such cases is straightforward: for the first term, 6 is multiplied by 5, the existing power of x, and the new power of x is 4 (as we subtract one from the original power). In essence, differentiation gives us a snapshot of the function's behavior at any given point, showing us how it's increasing or decreasing. This concept is vital not just for solving mathematical problems, but it also has practical applications in various sciences and engineering fields where understanding the rate of change is crucial.

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Most popular questions from this chapter

Runners stand at first and second base in a baseball game. At the moment a ball is hit, the runner at first base runs to second base at \(18 \mathrm{ft} / \mathrm{s} ;\) simultaneously, the runner on second runs to third base at \(20 \mathrm{ft} / \mathrm{s}\). How fast is the distance between the runners changing 1 second after the ball is hit (see figure)? (Hint: The distance between consecutive bases is \(90 \mathrm{ft}\) and the bases lie at the corners of a square.)

Suppose the position of an object moving horizontally after t seconds is given by the following functions \(s=f(t),\) where \(s\) is measured in feet, with \(s>0\) corresponding to positions right of the origin. a. Graph the position function. b. Find and graph the velocity function. When is the object stationary, moving to the right, and moving to the left? c. Determine the velocity and acceleration of the object at \(t=1\). d. Determine the acceleration of the object when its velocity is zero. e. On what intervals is the speed increasing? $$f(t)=2 t^{3}-21 t^{2}+60 t ; 0 \leq t \leq 6$$

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Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. For linear functions, the slope of any secant line always equals the slope of any tangent line. b. The slope of the secant line passing through the points \(P\) and \(Q\) is less than the slope of the tangent line at \(P\). c. Consider the graph of the parabola \(f(x)=x^{2} .\) For \(x > 0\) and \(h > 0,\) the secant line through \((x, f(x))\) and \((x+h, f(x+h))\) always has a greater slope than the tangent line at \((x, f(x))\)

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