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Calculate the derivative of the following functions. $$g(x)=\frac{x}{e^{3 x}}$$

Short Answer

Expert verified
Answer: The derivative of the function is $$g'(x) = \frac{1 - 3x}{e^{3x}}$$.

Step by step solution

01

Identify the functions f(x) and g(x) in the given function

In the given function, $$g(x) = \frac{x}{e^{3x}}$$, we can identify the functions f(x) and g(x) as: $$f(x) = x$$ $$g(x) = e^{3x}$$
02

Calculate the derivatives of f(x) and g(x)

Now, we will calculate the derivatives of both f(x) and g(x): For f(x) = x: $$f'(x) = 1$$ For g(x) = e^{3x} (using the Chain Rule with the derivative of e^u where u=3x): $$g'(x) = 3e^{3x}$$
03

Apply the Quotient Rule

Next, we will apply the Quotient Rule formula using the functions and their derivatives calculated in steps 1 and 2: $$\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$$ Substitute the values of f(x), f'(x), g(x), and g'(x): $$\frac{d}{dx}\left(\frac{x}{e^{3x}}\right) = \frac{1\cdot e^{3x} - x\cdot 3e^{3x}}{[e^{3x}]^2}$$
04

Simplify the expression

Now we will simplify the expression: $$\frac{d}{dx}\left(\frac{x}{e^{3x}}\right) = \frac{e^{3x} - 3xe^{3x}}{e^{6x}}$$ To further simplify the expression, factor out e^{3x} from the numerator: $$\frac{d}{dx}\left(\frac{x}{e^{3x}}\right) = \frac{e^{3x}(1 - 3x)}{e^{6x}}$$ Lastly, divide the numerator and denominator by e^{3x}: $$\frac{d}{dx}\left(\frac{x}{e^{3x}}\right) = \frac{1 - 3x}{e^{3x}}$$ The derivative of the given function, g(x), is: $$g'(x) = \frac{1 - 3x}{e^{3x}}$$

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Key Concepts

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

Quotient Rule
The quotient rule is a method for finding the derivative of a function that is the ratio of two differentiable functions. Imagine you have functions represented as \(f(x)\) being divided by \(g(x)\), like a fraction. The rule is expressed as: \[\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}\]
This formula is vital when dealing with fractions like in our exercise, where we have \(g(x)=\frac{x}{e^{3x}}\).
To utilize the quotient rule, you first need to differentiate the numerator \(f(x)\) and the denominator \(g(x)\) separately, as shown in the given solution. You then apply the derivatives according to the rule and simplify the expression. The quotient rule becomes extremely helpful in calculus when you encounter complex functions division.
Chain Rule
The chain rule is a principle in calculus used to find the derivative of a composite function. It is like unwinding layers, each representing a function. When you differentiate a composed function \(h(x) = f(g(x))\), the chain rule states: \[h'(x) = f'(g(x)) \cdot g'(x)\]
We use this method when we differentiate \(e^{3x}\) in our exercise because \(e^{3x}\) can be seen as an outer function \(f(u) = e^u\) with an inner function \(u=g(x) = 3x\). According to the chain rule, we multiply the derivative of the outer function with respect to the inner function, by the derivative of the inner function with respect to \(x\). By applying the chain rule correctly, we can effortlessly navigate through more complex derivations.
Exponential Functions
Exponential functions are mathematical expressions where the variable is an exponent - a classic example being \(e^{x}\). These functions are known for their unique property where the derivative is proportional to the function itself. For instance, the base of the natural exponential function, \(e\), is special because the derivative of \(e^{x}\) with respect to \(x\) is \(e^{x}\).
In our exercise, we encounter the exponential function \(e^{3x}\), which grows much faster than linear functions like \(f(x)=x\). Understanding the behavior and differentiation of exponential functions is crucial when dealing with growth and decay problems in various fields such as economics, biology, and physics. Simplifying derivatives of exponential functions typically involves using the chain rule along with the fundamental property of exponential functions.
Simplifying Expressions
Simplifying expressions is an essential skill in calculus, helping us present derivatives in their most compact and understandable form. After applying derivative rules, it is common to end up with complex, seemingly ungainly expressions. But fear not! With algebraic manipulation—like factoring, canceling, and reducing—we can streamline these expressions.
In the example given, after applying the quotient and chain rule, we ended up with \(\frac{e^{3x} - 3xe^{3x}}{e^{6x}}\). To simplify this, we factored out \(e^{3x}\) in the numerator and then reduced the expression by canceling out one \(e^{3x}\) from the numerator and denominator. Simplification is not just about aesthetics; it can reveal underlying properties and behavior of the function, making it a skill well worth mastering.

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Most popular questions from this chapter

Carry out the following steps. a. Verify that the given point lies on the curve. b. Determine an equation of the line tangent to the curve at the given point. $$x^{4}-x^{2} y+y^{4}=1 ;(-1,1)$$ (Graph cant copy)

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?

The total energy in megawatt-hr (MWh) used by a town is given by $$E(t)=400 t+\frac{2400}{\pi} \sin \frac{\pi t}{12}$$ where \(t \geq 0\) is measured in hours, with \(t=0\) corresponding to noon. a. Find the power, or rate of energy consumption, \(P(t)=E^{\prime}(t)\) in units of megawatts (MW). b. At what time of day is the rate of energy consumption a maximum? What is the power at that time of day? c. At what time of day is the rate of energy consumption a minimum? What is the power at that time of day? d. Sketch a graph of the power function reflecting the times when energy use is a minimum or a maximum.

Earth's atmospheric pressure decreases with altitude from a sea level pressure of 1000 millibars (a unit of pressure used by meteorologists). Letting \(z\) be the height above Earth's surface (sea level) in kilometers, the atmospheric pressure is modeled by \(p(z)=1000 e^{-z / 10}\) a. Compute the pressure at the summit of Mt. Everest, which has an elevation of roughly \(10 \mathrm{km} .\) Compare the pressure on Mt. Everest to the pressure at sea level. b. Compute the average change in pressure in the first \(5 \mathrm{km}\) above Earth's surface. c. Compute the rate of change of the pressure at an elevation of \(5 \mathrm{km}\) d. Does \(p^{\prime}(z)\) increase or decrease with \(z ?\) Explain. e. What is the meaning of \(\lim _{z \rightarrow \infty} p(z)=0 ?\)

Use implicit differentiation to find\(\frac{d y}{d x}.\) $$y=\frac{x+1}{y-1}$$

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