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Derivatives Find the derivative of the following functions. See Example 2 of Section 3.2 for the derivative of \(\sqrt{x}\). $$g(w)=\frac{5}{6} w^{12}$$

Short Answer

Expert verified
Answer: The derivative of the function is \(g'(w) = 2w^{11}\).

Step by step solution

01

Identify the function

We are given the function \(g(w) = \frac{5}{6} w^{12}\) and we need to find its derivative.
02

Use the power rule

Applying the power rule to the function \(g(w) = \frac{5}{6} w^{12}\), we get: $$g'(w) = \frac{5}{6} (12w^{(12-1)})$$
03

Simplify the derivative

Now, we'll simplify the expression: $$g'(w) = \frac{5}{6} (12w^{11})$$ \(g'(w) = 2w^{11}\) The derivative of the function \(g(w) = \frac{5}{6} w^{12}\) is \(g'(w) = 2w^{11}\).

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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 tool in calculus used for finding the derivatives of polynomial functions. It states that for a function of the form \(f(x) = x^n\), its derivative \(f'(x)\) is \(nx^{n-1}\). This rule simplifies the process of differentiation, especially for students beginning their journey into calculus. In the given exercise, we apply the power rule to the function \(g(w) = \frac{5}{6} w^{12}\). Here, the term \(w^{12}\) follows the format necessitated by the power rule, where \(n = 12\). By applying the rule, the exponent \(12\) is brought down in front of the term, and the new exponent is decremented by one, producing \(12w^{11}\). The constant, \(\frac{5}{6}\), remains unaffected by the power differentiation and is multiplied by the result. This is how the power rule streamlines finding derivatives.
Differentiation
Differentiation is the process of finding the derivative of a function. A derivative represents the rate at which a function is changing at any given point and is fundamental to calculus. In practical terms, when you differentiate a function, you are finding its slope or gradient. This helps in understanding how a function behaves over an interval. The process involves systematic rules, one of which is the power rule. Differentiation becomes particularly useful in real-world problems where changes need to be tracked, such as in physics to compute velocity or in economics to assess change in profit margins. In our example, \(g(w) = \frac{5}{6} w^{12}\), differentiation via the power rule gives us \(g'(w) = 2w^{11}\), unveiling how the function varies with change in \(w\). This understanding is key to tackling more complex calculus problems.
Calculus Fundamentals
Calculus is the branch of mathematics that studies continuous change. It is divided primarily into differentiation and integration. Its principles are applied across a broad spectrum, including science, engineering, and economics, making it a critical area of study. At its heart, calculus enables us to model dynamic systems and understand their behavior. Key fundamentals include limits, functions, and rates of change. Specifically, derivatives—like the one we calculated for \(g(w) = \frac{5}{6} w^{12}\)—are fundamental tools that provide a snapshot of the function's rate of change at any point. Grasping these basics helps students transition smoothly into more advanced topics where these elements and processes are applied to multidimensional problems further showcasing calculus's power and utility.

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Most popular questions from this chapter

Beginning at age \(30,\) a self-employed plumber saves \(\$ 250\) per month in a retirement account until he reaches age \(65 .\) The account offers \(6 \%\) interest, compounded monthly. The balance in the account after \(t\) years is given by \(A(t)=50,000\left(1.005^{12 t}-1\right)\) a. Compute the balance in the account after \(5,15,25,\) and 35 years. What is the average rate of change in the value of the account over the intervals \([5,15],[15,25],\) and [25,35]\(?\) b. Suppose the plumber started saving at age 25 instead of age 30\. Find the balance at age 65 (after 40 years of investing). c. Use the derivative \(d A / d t\) to explain the surprising result in part (b) and the advice: Start saving for retirement as early as possible.

\- Tangency question It is easily verified that the graphs of \(y=1.1^{x}\) and \(y=x\) have two points of intersection, and the graphs of \(y=2^{x}\) and \(y=x\) have no point of intersection. It follows that for some real number \(1.1 < p < 2,\) the graphs of \(y=p^{x}\) and \(y=x\) have exactly one point of intersection. Using analytical and/or graphical methods, determine \(p\) and the coordinates of the single point of intersection.

The number of hours of daylight at any point on Earth fluctuates throughout the year. In the Northern Hemisphere, the shortest day is on the winter solstice and the longest day is on the summer solstice. At \(40^{\circ}\) north latitude, the length of a day is approximated by $$D(t)=12-3 \cos \left(\frac{2 \pi(t+10)}{365}\right)$$ where \(D\) is measured in hours and \(0 \leq t \leq 365\) is measured in days, with \(t=0\) corresponding to January 1 a. Approximately how much daylight is there on March 1 \((t=59) ?\) b. Find the rate at which the daylight function changes. c. Find the rate at which the daylight function changes on March \(1 .\) Convert your answer to units of min/day and explain what this result means. d. Graph the function \(y=D^{\prime}(t)\) using a graphing utility. e. At what times of the year is the length of day changing most rapidly? Least rapidly?

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 that they actually lie on the curve. Confirm your results with a graph. $$x^{2}\left(3 y^{2}-2 y^{3}\right)=4$$

The output of an economic system \(Q,\) subject to two inputs, such as labor \(L\) and capital \(K\) is often modeled by the Cobb-Douglas production function \(Q=c L^{a} K^{b} .\) When \(a+b=1,\) the case is called constant returns to scale. Suppose \(Q=1280, a=\frac{1}{3}, b=\frac{2}{3},\) and \(c=40.\) a. Find the rate of change of capital with respect to labor, \(d K / d L.\) b. Evaluate the derivative in part (a) with \(L=8\) and \(K=64.\)

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