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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

Work carefully Proceed with caution when using implicit differentiation to find points at which a curve has a specified slope. For the following curves, find the points on the curve (if they exist) at which the tangent line is horizontal or vertical. Once you have found possible points, make sure they actually lie on the curve. Confirm your results with a graph. $$x\left(1-y^{2}\right)+y^{3}=0$$

General logarithmic and exponential derivatives Compute the following derivatives. Use logarithmic differentiation where appropriate. $$\frac{d}{d x}\left(x^{\cos x}\right)$$

Calculating limits exactly Use the definition of the derivative to evaluate the following limits. $$\lim _{h \rightarrow 0} \frac{\ln \left(e^{8}+h\right)-8}{h}$$

Suppose a large company makes 25,000 gadgets per year in batches of \(x\) items at a time. After analyzing setup costs to produce each batch and taking into account storage costs, it has been determined that the total cost \(C(x)\) of producing 25,000 gadgets in batches of \(x\) items at a time is given by $$C(x)=1,250,000+\frac{125,000,000}{x}+1.5 x.$$ a. Determine the marginal cost and average cost functions. Graph and interpret these functions. b. Determine the average cost and marginal cost when \(x=5000\). c. The meaning of average cost and marginal cost here is different from earlier examples and exercises. Interpret the meaning of your answer in part (b).

Jean and Juan run a one-lap race on a circular track. Their angular positions on the track during the race are given by the functions \(\theta(t)\) and \(\varphi(t),\) respectively, where \(0 \leq t \leq 4\) and \(t\) is measured in minutes (see figure). These angles are measured in radians, where \(\theta=\varphi=0\) represent the starting position and \(\theta=\varphi=2 \pi\) represent the finish position. The angular velocities of the runners are \(\theta^{\prime}(t)\) and \(\varphi^{\prime}(t)\). a. Compare in words the angular velocity of the two runners and the progress of the race. b. Which runner has the greater average angular velocity? c. Who wins the race? d. Jean's position is given by \(\theta(t)=\pi t^{2} / 8 .\) What is her angular velocity at \(t=2\) and at what time is her angular velocity the greatest? e. Juan's position is given by \(\varphi(t)=\pi t(8-t) / 8 .\) What is his angular velocity at \(t=2\) and at what time is his angular velocity the greatest?

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