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Derivatives Find the derivative of the following functions. See Example 2 of Section 3.2 for the derivative of \(\sqrt{x}\). $$g(x)=e^{3}$$

Short Answer

Expert verified
Answer: The derivative of the function \(g(x) = e^3\) with respect to x is 0.

Step by step solution

01

Identify the type of function

From the given function, we can see that it's a constant function. The constant function \(g(x) = e^3\) does not depend on the variable \(x\).
02

Apply the constant rule of differentiation

The constant rule of differentiation states that the derivative of any constant function is zero. Thus, the derivative of \(g(x) = e^3\) with respect to x is 0.
03

Write the final answer

Since the derivative of \(g(x) = e^3\) with respect to x is 0, we can write the answer as: $$ \frac{d}{dx}g(x) = 0 $$

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Key Concepts

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

Constant Rule of Differentiation
When tackling calculus problems, one often encounters functions that seem complex at first glance. A common type of function you'll meet is the constant function. But no need to worry, the constant rule of differentiation is here to simplify things! This rule is key for understanding how to deal with these constant functions.

Imagine you're observing something that does not change with time, like the timeless beauty of ancient architecture. This lack of change is analogous to a constant function in mathematics—in both cases, there's no variation. The constant rule of differentiation taps into this idea. It states that if you have a function that always outputs the same number, no matter the input, its derivative is always zero. This is logical because if you're not moving, your speed—or rate of change—is zero, right?

In our exercise example, the function is a constant value, which is Euler's number raised to the power of three, or simply, \(e^3\). According to the rule, no matter what value of \(x\) you plug in, the output remains steadfastly at \(e^3\). Thus, when you differentiate it, you are essentially asking, 'How does the function change as \(x\) changes?' The answer: it doesn't change at all, therefore the derivative is zero.
Calculus

The Big Picture of Calculus

In the grand scheme of mathematics, calculus holds a special place. It's the study of how things change, giving us the tools to model and control systems, predict phenomena, and understand the dynamics behind motion and growth. When you're working through calculus problems, try to visualize it as the mathematics of motion and change, like a language describing the movements in nature.

Calculus is commonly divided into two main branches: differential calculus and integral calculus. Differential calculus, which includes our constant rule of differentiation, deals with the rate of change of quantities—the derivatives, while integral calculus focuses on the aggregation of quantities—the antiderivatives or integrals. Both branches are connected by the fundamental theorem of calculus, bridging the gap between them and providing a powerful framework for solving problems related to change and area under curves.
Derivative
At its heart, the derivative is a concept from calculus that measures the sensitivity to change of a quantity, which is usually a function of some other quantity. Think of it as an instantaneous snapshot of how one thing is affected as the other changes—similar to checking the speedometer in a car for your immediate speed, not your average speed over a trip.

Mathematically, the derivative of a function at a point is the slope of the tangent to the function's curve at that point. It's defined as the limit of the ratio of the change of the function value to the change in the input value as the change in input value becomes infinitely small. In tangible terms, it tells you how steeply a function is climbing or falling at a particular point on the curve. When students approach problems about derivatives, understanding this fundamental notion—that derivatives represent a rate of change—is key to grasping the more abstract applications and operations within calculus.

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Most popular questions from this chapter

Use implicit differentiation to find\(\frac{d y}{d x}.\) $$x+2 y=\sqrt{y}$$

The population of a culture of cells after \(t\) days is approximated by the function \(P(t)=\frac{1600}{1+7 e^{-0.02 t}},\) for \(t \geq 0\) a. Graph the population function. b. What is the average growth rate during the first 10 days? c. Looking at the graph, when does the growth rate appear to be a maximum? d. Differentiate the population function to determine the growth rate function \(P^{\prime}(t)\) e. Graph the growth rate. When is it a maximum and what is the population at the time that the growth rate is a maximum?

a. Determine an equation of the tangent line and the normal line at the given point \(\left(x_{0}, y_{0}\right)\) on the following curves. (See instructions for Exercises 73-78. b. Graph the tangent and normal lines on the given graph. \(\left(x^{2}+y^{2}\right)^{2}=\frac{25}{3}\left(x^{2}-y^{2}\right); \left(x_{0}, y_{0}\right)=(2,-1)\) (lemniscate of Bernoulli) (Graph cant copy)

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.

Carry out the following steps. a. Use implicit differentiation to find \(\frac{d y}{d x}\). b. Find the slope of the curve at the given point. $$\sqrt[3]{x}+\sqrt[3]{y^{4}}=2 ;(1,1)$$

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