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Find the derivative of each function. $$ f(x)=x \ln x-x $$

Short Answer

Expert verified
The derivative is \( f'(x) = \ln x \).

Step by step solution

01

Identify the components of the function

The function given is \( f(x) = x \ln x - x \). It has two main components: \( x \ln x \) and \( -x \). We will find the derivative of each term separately and then combine them.
02

Differentiate the first term \( x \ln x \)

To find the derivative of \( x \ln x \), we use the product rule. Recall the product rule is \( (u \, v)' = u'v + uv' \). Here, \( u = x \) and \( v = \ln x \).The derivative \( u' = 1 \) and \( v' = \frac{1}{x} \).Applying the product rule, we get:\[(x \ln x)' = (1)(\ln x) + (x)\left(\frac{1}{x}\right) = \ln x + 1\]
03

Differentiate the second term \(-x\)

The derivative of \(-x\) is straightforward. It is simply \(-1\).
04

Combine the derivatives

Now, combine the derivatives found in the previous steps. The derivative of \( x \ln x \) is \( \ln x + 1 \) and the derivative of \(-x\) is \(-1\). Therefore, the derivative of the entire function is:\[f'(x) = (\ln x + 1) + (-1)\]This simplifies to:\[f'(x) = \ln x\]
05

Conclusion

The derivative of the function \( f(x) = x \ln x - x \) with respect to \( x \) is \( f'(x) = \ln x \). This means the rate of change of the function, or the slope of the tangent line to the curve at any point \( x \), is \( \ln x \).

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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 used to differentiate functions that are products of two or more simpler functions. This rule simplifies the process of finding derivatives of composite functions. It states that if you have a product of two functions, say \( u(x) \) and \( v(x) \), the derivative of their product is given by:
  • \[ (uv)' = u'v + uv' \]
This formula allows us to break down the task into finding the derivatives of the individual components \( u(x) \) and \( v(x) \), and then applying them together in the formula above.
In the example exercise, the term \( x \ln x \) is such a product where \( u = x \) and \( v = \ln x \). Here,
  • \( u' = 1 \) because the derivative of \( x \) is \( 1 \).
  • \( v' = \frac{1}{x} \) because the derivative of \( \ln x \) is \( \frac{1}{x} \).
By applying the product rule formula, we can find the derivative of \( x \ln x \) as: \( (x \ln x)' = \ln x + 1 \). This method is essential because it allows us to handle more complex functions found in calculus.
Logarithmic Differentiation
Logarithmic differentiation is a powerful technique, especially useful when the function involves products, quotients, or powers of functions. It involves taking the natural logarithm of both sides of an equation and then differentiating using the properties of logarithms.
Using logarithmic differentiation can simplify the process by turning multiplicative processes into additive, thanks to the rule that \( \ln(ab) = \ln a + \ln b \). While it wasn’t directly used in this solution, it is an important technique to understand when dealing with complex functions involving logarithms.
In cases where the function is in the form of \( x^x \) or similar, logarithmic differentiation makes things easier by simplifying exponentials and products for easier differentiation. To apply, set \( y = f(x) \) and take the natural log on both sides so it becomes \( \ln y = \ln(f(x)) \). Differentiate both sides, using the chain rule on \( y \), to solve for \( y' \).
This approach is often used in conjunction with other rules, like the product rule, for maximum effectiveness.
Differentiation Techniques
Differentiation techniques in calculus are essential for finding the rate of change of functions. They provide a range of methods for handling different forms and complexities of functions. Common techniques include:
  • Power Rule: Easily finds the derivative of any power of \( x \), \( \frac{d}{dx}(x^n) = nx^{n-1} \).
  • Product Rule: As previously discussed, helpful for differentiating products of functions.
  • Quotient Rule: Useful for differentiating quotients, \( \frac{u}{v} \), where \( (\frac{u}{v})' = \frac{u'v - uv'}{v^2} \).
  • Chain Rule: Necessary when differentiating composite functions, allowing us to work from the outermost to the innermost function layers.
Each technique has a specific use-case that makes it effective under certain conditions. By understanding and recognizing which rule to apply, finding the derivatives of more intricate functions becomes manageable.
In our exercise, the "simpler" rules like the product rule were used to break down the work into manageable steps, demonstrating their value in practical applications.

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Most popular questions from this chapter

A supply function \(S(p)\) gives the total amount of a product that producers are willing to supply at a given price \(p\). The elasticity of supply is defined as $$ E_{s}(p)=\frac{p \cdot S^{\prime}(p)}{S(p)} $$ Elasticity of supply measures the relative increase in supply resulting from a small relative increase in price. It is less useful than elasticity of demand, however, since it is not related to total revenue. Use the preceding formula to find the elasticity of supply for a supply function of the form \(S(p)=a p^{n}\), where \(a\) and \(n\) are positive constants.

Use your graphing calculator to graph each function on a window that includes all relative extreme points and inflection points, and give the coordinates of these points (rounded to two decimal places). [Hint: Use NDERIV once or twice with ZERO.] (Answers may vary depending on the graphing window chosen.) $$ f(x)=1-e^{-x^{2} / 2} $$

A sum of $$\$ 1000$$ at \(5 \%\) interest compounded continuously will grow to \(V(t)=1000 e^{0.05 t}\) dollars in \(t\) years. Find the rate of growth after: a. 0 years (the time of the original deposit). b. 10 years.

If demand is elastic and you lower prices, would you expect demand to increase or decrease, and strongly or weakly?

Use your graphing calculator to graph each function on a window that includes all relative extreme points and inflection points, and give the coordinates of these points (rounded to two decimal places). [Hint: Use NDERIV once or twice with ZERO.] (Answers may vary depending on the graphing window chosen.) $$ f(x)=e^{x}+e^{-x} $$

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