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Find the derivative of each function. \(f(x)=\left(x+2 e^{x}\right)\left(x-e^{x}\right)\)

Short Answer

Expert verified
The derivative of the function \(f(x) = (x + 2e^x)(x - e^x)\) is: \(f'(x) = 2x - e^x + 2xe^x - xe^x - 4e^{2x}\)

Step by step solution

01

Find the derivative using the product rule

The product rule states that if we have a function in the form of \(g(x) = u(x) \cdot v(x)\), then the derivative is given by \(g'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)\). Here, we can think of \(u(x) = x + 2e^x\) and \(v(x) = x - e^x\).
02

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

First, we need to find the derivatives of \(u(x)\) and \(v(x)\). We use the chain rule: For \(u(x)\): \(u'(x) = \frac{d}{dx}(x + 2e^x) = 1 + 2 \cdot e^x\) For \(v(x)\): \(v'(x) = \frac{d}{dx}(x - e^x) = 1 - e^x\)
03

Apply the product rule

Now we can apply the product rule: \(f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)\) Substitute the values we found in the previous steps: \(f'(x) = (1 + 2e^x)(x - e^x) + (x + 2e^x)(1 - e^x)\)
04

Expand and simplify

Now, we will expand the expression and simplify the terms: \(f'(x) = (x - e^x) + 2e^x(x - e^x) + (x + 2e^x) - (x + 2e^x)e^x\) Distribute the terms: \(f'(x) = x - e^x + 2xe^x - 2e^{2x} + x + 2e^x - xe^x - 2e^{2x}\) Combine like terms: \(f'(x) = 2x - e^x + 2xe^x - xe^x - 4e^{2x}\)
05

Final result

Thus, the derivative of the function \(f(x) = (x + 2e^x)(x - e^x)\) is: \(f'(x) = 2x - e^x + 2xe^x - xe^x - 4e^{2x}\)

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Key Concepts

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

Product Rule
The product rule is a fundamental technique in calculus used to find the derivative of products of two functions. When given a function that is the product of two sub-functions, say, \( u(x) \) and \( v(x) \), the product rule enables us to differentiate this product by applying a specific formula: \( (uv)' = u'v + uv' \).

In the context of the exercise, \( u(x) = x + 2e^x \) and \( v(x) = x - e^x \). Following the product rule, we find the derivatives \( u'(x) \) and \( v'(x) \) first before applying the rule itself to obtain the derivative of the entire function.
Chain Rule
The chain rule is essential when dealing with composite functions, in which one function is nested inside another. In its simple form, the chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. Mathematically, if \( h(x) = f(g(x)) \), then \( h'(x) = f'(g(x))g'(x) \).

The exercise presented involves exponential functions within other expressions, which necessitates the use of the chain rule for differentiation. While the chain rule isn't directly highlighted in the steps, it's implicitly used in finding the derivatives of \( u(x) \) and \( v(x) \), particularly in differentiating terms involving \( e^x \), an exponential function.
Exponential Functions
Exponential functions, typically denoted as \( e^x \), have unique properties in calculus. They are special because their rate of growth is directly proportional to their value, which becomes evident in their derivative, which is also \( e^x \).

In the given exercise, we encounter exponential functions within the product of two functions. Deriving terms that contain \( e^x \) will not change the base function because the derivative of \( e^x \) is \( e^x \) itself. This property simplifies the differentiation process and is a key element when expanding and simplifying the final expression in the exercise.
Simplifying Derivatives
Simplifying derivatives helps in transforming complex expressions into more manageable forms. The process typically involves expanding products, combining like terms, and canceling where appropriate to achieve the simplest expression of the derivative.

In our exercise, once we apply the product rule, the next step is to expand the products and combine like terms to simplify the derivative. This step ensures clarity and provides a straightforward representation of the derivative. The final expression, after careful simplification, encompasses the essence of the original function's rate of change with respect to \( x \).

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Most popular questions from this chapter

Plot the graph the function and use it to guess at the limit. Verify your result using l'Hôpital's Rule. $$ \lim _{x \rightarrow 1}\left(\frac{1}{\ln x}-\frac{1}{x-1}\right) $$

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In Exercises, (a) find the equations of the tangent and the normal lines to the curve at the indicated point. (The normal line at a point on the curve is the line perpendicular to the tangent line at that point.) (b) Then use a graphing utility to plot the curve and the tangent and normal lines on the same screen. $$ 4 x y-9=0 ; \quad\left(3, \frac{3}{4}\right) $$

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