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Use the General Power Rule where appropriate to find the derivative of the following functions. $$f(x)=\frac{2^{x}}{2^{x}+1}$$

Short Answer

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Question: Find the derivative of the function $$f(x)=\frac{2^{x}}{2^{x}+1}$$. Answer: The derivative of the given function is: $$f'(x) = \frac{2^{x}\ln(2)}{(2^{x}+1)}-\frac{2^{x}\ln^2(2)}{(2^{x}+1)}$$.

Step by step solution

01

Rewrite the function in the form of $$f(x)=u(x)^{v(x)}$$

We can rewrite the given function $$f(x)=\frac{2^{x}}{2^{x}+1}$$ as: $$f(x)=(2^{x}(2^{x}+1)^{-1})$$ Now, we have $$u(x) = 2^{x}$$ and $$v(x) = (2^{x}+1)^{-1}$$. Step 2: Find the derivatives of $$u(x)$$ and $$v(x)$$
02

Calculate $$u'(x)$$ and $$v'(x)$$

Derivative of $$u(x)=2^{x}$$ is: $$u'(x)=2^{x}\ln(2)$$ Derivative of $$v(x)=(2^{x}+1)^{-1}$$ can be found using the Chain Rule: $$v'(x)=\frac{d}{dx}(2^{x}+1)^{-1} = -1(2^{x}+1)^{-2} \cdot \frac{d}{dx}(2^{x}+1) = -1(2^{x}+1)^{-2} \cdot 2^{x}\ln(2) = -\frac{2^{x}\ln(2)}{(2^{x}+1)^2}$$ Step 3: Apply the General Power Rule
03

Apply the General Power Rule to find $$f'(x)$$

Now we will use the General Power Rule to find the derivative of the given function: $$f'(x) = u'(x)v(x)u(x)^{v(x)-1} + v'(x)u(x)^{v(x)}\ln(u(x))$$ Substituting the values we have: $$f'(x) = (2^{x}\ln(2))(2^{x}+1)^{-1}(2^{x}(2^{x}+1)^{-1})^{(2^{x}+1)^{-1}-1} - \frac{2^{x}\ln(2)}{(2^{x}+1)^2}(2^{x}(2^{x}+1)^{-1})^{(2^{x}+1)^{-1}}\ln(2^{x})$$ Now, simplify the expression: $$f'(x) = (2^{x}\ln(2))(2^{x}+1)^{-1} - \frac{2^{x}\ln^2(2)}{(2^{x}+1)}$$ So, the derivative of the given function is: $$f'(x) = \frac{2^{x}\ln(2)}{(2^{x}+1)}-\frac{2^{x}\ln^2(2)}{(2^{x}+1)}$$

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Key Concepts

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

Chain Rule
The Chain Rule is a fundamental concept in calculus used for finding the derivative of a composition of functions. When you have a function nested within another function, the Chain Rule helps to differentiate them step by step.

To apply the Chain Rule, follow these steps:
  • Differently identify the inner function and the outer function.
  • Find the derivative of the outer function, leaving the inner function unchanged.
  • Multiply this derivative by the derivative of the inner function.
In the exercise above, we used the Chain Rule to find the derivative of \(v(x) = (2^{x} + 1)^{-1}\). Here, the outer function is \(g(y) = y^{-1}\) and the inner function is \(h(x) = 2^{x} + 1\).

First, we differentiated the outer function while keeping the inner function intact, giving us \(-1(2^{x} + 1)^{-2}\). We then multiplied this by the derivative of the inner function, \(2^{x} \ln(2)\), resulting in the final derivative \(-\frac{2^{x}\ln(2)}{(2^{x}+1)^2}\). Giving the systematic method of combining the derivatives of nested functions, the Chain Rule becomes immensely useful, especially when dealing with complex compositions.
Derivatives
Derivatives represent the rate at which a function changes. In simple terms, finding a derivative means finding the slope of a function at any given point.

Why do derivatives matter?
  • They help determine the behavior of functions.
  • They're fundamental in finding tangents and setting up function optimization problems.
  • In real-world applications, they describe how one quantity changes in response to another.
In the context of the exercise, we focused on finding derivatives for parts of the function \(f(x) = \frac{2^{x}}{2^{x} + 1}\).

The derivative of \(u(x) = 2^{x}\) was calculated as \(2^{x}\ln(2)\), using the rules of exponential differentiation. Additionally, for \(v(x) = (2^{x} + 1)^{-1}\), we used the Chain Rule to derive \(-\frac{2^{x}\ln(2)}{(2^{x}+1)^2}\).

In finding the complete derivative \(f'(x)\), we apply steps using both the General Power Rule and the Chain Rule to systematically find the rate of change for the given function, reflecting the complexity and beauty of derivatives working in concert.
Exponential Functions
Exponential functions are functions of the form \(a^{x}\) where \(a\) is a constant and \(x\) is the exponent. These functions are crucial in modeling growth and decay processes.

Key Properties of Exponential Functions
  • Their rates of change grow proportionally with the current value, leading to fast growth or decay.
  • They have continuous and smooth curves on graphs.
  • Common in scientific and financial calculations due to natural growth processes.
In mathematics, differentiation of exponential functions involves a neat formula. If \(f(x) = a^{x}\), then the derivative \(f'(x) = a^{x} \ln(a)\).

In the exercise, differentiation of the exponential function \(2^{x}\) used this rule to obtain \(2^{x}\ln(2)\). Exponential functions, particularly when paired with powers or nested within other functions, often require the use of advanced calculus methods like the Chain Rule and General Power Rule to differentiate, as we did in the exercise when working through \(f(x) = \frac{2^{x}}{2^{x} + 1}\). Understanding and applying these rules allow for adeptly managing the dynamics of exponential expressions in calculus.

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Most popular questions from this chapter

An observer stands \(20 \mathrm{m}\) from the bottom of a 10 -m-tall Ferris wheel on a line that is perpendicular to the face of the Ferris wheel. The wheel revolves at a rate of \(\pi \mathrm{rad} / \mathrm{min},\) and the observer's line of sight with a specific seat on the wheel makes an angle \(\theta\) with the ground (see figure). Forty seconds after that seat leaves the lowest point on the wheel, what is the rate of change of \(\theta ?\) Assume the observer's eyes are level with the bottom of the wheel.

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Identifying functions from an equation. The following equations implicitly define one or more functions. a. Find \(\frac{d y}{d x}\) using implicit differentiation. b. Solve the given equation for \(y\) to identify the implicitly defined functions \(y=f_{1}(x), y=f_{2}(x), \ldots\) c. Use the functions found in part (b) to graph the given equation. $$x^{4}=2\left(x^{2}-y^{2}\right) \text { (eight curve) }$$

A thin copper rod, 4 meters in length, is heated at its midpoint, and the ends are held at a constant temperature of \(0^{\circ} .\) When the temperature reaches equilibrium, the temperature profile is given by \(T(x)=40 x(4-x),\) where \(0 \leq x \leq 4\) is the position along the rod. The heat flux at a point on the rod equals \(-k T^{\prime}(x),\) where \(k>0\) is a constant. If the heat flux is positive at a point, heat moves in the positive \(x\) -direction at that point, and if the heat flux is negative, heat moves in the negative \(x\) -direction. a. With \(k=1,\) what is the heat flux at \(x=1 ?\) At \(x=3 ?\) b. For what values of \(x\) is the heat flux negative? Positive? c. Explain the statement that heat flows out of the rod at its ends.

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