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Find the derivative of the following functions. \(f(x)=10 x^{4}-32 x+e^{2}\)

Short Answer

Expert verified
Answer: The derivative of the function $$f(x) = \frac{e^x}{e^x + 1}$$ is $$f'(x) = \frac{e^x}{(e^x + 1)^2}$$.

Step by step solution

01

Identify the derivative rule to use

Since the given function is a fraction, we will use the quotient rule to find the derivative. The quotient rule states that given a function $$h(x) = \frac{f(x)}{g(x)}$$, its derivative is given by: $$h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$$ Our function is $$f(x)=\frac{e^{x}}{e^{x}+1}$$, so we have: $$f(x) = \frac{e^{x}}{e^{x}+1} = \frac{u(x)}{v(x)}$$ Where $$u(x) = e^{x}$$ and $$v(x) = e^{x} + 1$$
02

Find the derivatives of u(x) and v(x)

We need to find $$u'(x)$$ and $$v'(x)$$ (the first derivatives of u(x) and v(x)): The derivative of $$u(x) = e^{x}$$ is: $$u'(x) = \frac{d}{dx} e^{x} = e^{x}$$ The derivative of $$v(x) = e^{x} + 1$$ is: $$v'(x) = \frac{d}{dx} (e^{x} + 1) = e^{x}$$
03

Apply the quotient rule

Using the quotient rule and the derivatives we just found, we can now compute the derivative of the function: $$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$$ Plug in the values for $$u'(x)$$, $$v(x)$$, $$u(x)$$, and $$v'(x)$$: $$f'(x) = \frac{e^{x}(e^{x}+1) - e^{x}e^{x}}{(e^{x}+1)^2}$$
04

Simplify the expression

Simplify the numerator and the final expression for the derivative: $$f'(x) = \frac{e^{2x}+e^{x} - e^{2x}}{(e^{x}+1)^2} = \frac{e^{x}}{(e^{x}+1)^2}$$ So the derivative of $$f(x)=\frac{e^{x}}{e^{x}+1}$$ is: $$f'(x) = \frac{e^{x}}{(e^{x}+1)^2}$$

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Key Concepts

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

Exponential Functions
Exponential functions are a key player in calculus, especially when it comes to solving derivatives. An exponential function has the form \(a^{x}\), where \(a\) is a positive constant, and \(x\) is the exponent. In these problems, we frequently encounter the exponential function \(e^{x}\), where \(e\) is approximately 2.718, the base of natural logarithms. Exponential functions grow rapidly, and their unique property is that they are unchanged by differentiation.

This means when you take the derivative of \(e^{x}\), it remains \(e^{x}\). This property greatly simplifies finding derivatives, especially in complex functions involving exponential expressions. Understanding how to manipulate and differentiate exponential functions is foundational, and recognizing their patterns can make calculus much more approachable.
Quotient Rule
The Quotient Rule is an essential differentiation technique when dealing with functions that are represented as one function divided by another. This is often the case with rational functions or fractions like \(\frac{e^x}{e^x + 1}\).

The rule itself is a formula that allows us to differentiate these kinds of functions by taking the derivatives of the numerator and the denominator separately. The quotient rule is expressed as:
  • If \(h(x) = \frac{f(x)}{g(x)}\), then \(h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}\).
By applying this rule, we methodically find the derivatives of the respective parts of the function. For example, in the function \(\frac{e^x}{e^x + 1}\), we differentiate \(e^x\) and \(e^x + 1\) as separate entities. After finding these derivatives, they are inserted into the quotient rule formula to yield the complete derivative of the original function.

It might seem complex at first, but with practice, the process becomes more intuitive. The key is to keep the format of the rule in mind and apply it patiently.
Differentiation Techniques
Differentiation techniques encompass various rules and methods used for determining derivatives in calculus. These techniques are the backbone of many calculus problems. Three fundamental rules often used are the power rule, product rule, and quotient rule.

The exercise we discussed earlier uses the quotient rule, which is applied specifically for functions that are arranged as fractions. The power rule helps with polynomial expressions, and the product rule applies when two functions are multiplied. Each rule comes with its formula and way of application. Mastery of these techniques allows for solving complex calculus problems efficiently.

Consistent practice with these methods will lead to a better understanding and ability to tackle different scenarios of differentiation. As students become more familiar, they will start recognizing which technique to apply in any given situation, which is essential for problem-solving and deeper comprehension of calculus.

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Most popular questions from this chapter

Proof of the Quotient Rule Let \(F=f / g\) be the quotient of two functions that are differentiable at \(x\) a. Use the definition of \(F^{\prime}\) to show that \(\frac{d}{d x}\left[\frac{f(x)}{g(x)}\right]=\lim _{h \rightarrow 0} \frac{f(x+h) g(x)-f(x) g(x+h)}{h g(x+h) g(x)}\) b. Now add \(-f(x) g(x)+f(x) g(x)\) (which equals 0) to the numerator in the preceding limit to obtain $$\lim _{h \rightarrow 0} \frac{f(x+h) g(x)-f(x) g(x)+f(x) g(x)-f(x) g(x+h)}{h g(x+h) g(x)}$$ Use this limit to obtain the Quotient Rule. c. Explain why \(F^{\prime}=(f / g)^{\prime}\) exists, whenever \(g(x) \neq 0\)

Orthogonal trajectories Two curves are orthogonal to each other if their tangent lines are perpendicular at each point of intersection (recall that two lines are perpendicular to each other if their slopes are negative reciprocals. . A family of curves forms orthogonal trajectories with another family of curves if each curve in one family is orthogonal to each curve in the other family. For example, the parabolas \(y=c x^{2}\) form orthogonal trajectories with the family of ellipses \(x^{2}+2 y^{2}=k,\) where \(c\) and \(k\) are constants (see figure). Use implicit differentiation if needed to find \(d y / d x\) for each equation of the following pairs. Use the derivatives to explain why the families of curves form orthogonal trajectories. \(y=c x^{2} ; x^{2}+2 y^{2}=k,\) where \(c\) and \(k\) are constants

Find the derivative of the inverse of the following functions at the specified point on the graph of the inverse function. You do not need to find \(f^{-1}\) $$f(x)=x^{2}+1, \text { for } x \geq 0 ;(5,2)$$

Find \(f^{\prime}(x), f^{\prime \prime}(x),\) and \(f^{\prime \prime \prime}(x)\) \(f(x)=\frac{x^{2}-7 x}{x+1}\)

Find the slope of the curve \(5 \sqrt{x}-10 \sqrt{y}=\sin x\) at the point \((4 \pi, \pi)\).

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