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Chapter 2: Problem 33

Use logarithmic differentiation to find the derivative of the function. Find \(y^{\prime \prime}\) if \(y=x^{x}\).

Short Answer

Expert verified
The second derivative of the given function, \(y = x^x\), is: \( y'' = x^{x-1} \ln(x) + x^{x-1} + x^x(\ln(x) + 1)^2 \).

Step by step solution

01

Apply natural logarithm to both sides

To be able to perform logarithmic differentiation, we first take the natural logarithm of both sides of the equation: \( \ln(y) = \ln(x^x) \)
02

Use properties of logarithms

Using the properties of logarithms, we can simplify the right-hand side of the equation: \( \ln(y) = x \ln(x) \)
03

Differentiate with respect to x

Now we need to differentiate both sides of the equation with respect to x. By doing so, we get: \( \frac{1}{y}\frac{dy}{dx} = \frac{d}{dx}(x \ln(x)) \)
04

Use Product Rule to find the first derivative, y'

Using the product rule, we can find the first derivative: \( \frac{1}{y}\frac{dy}{dx} = (1) \ln(x) + x \frac{1}{x} \) Simplifying, we get: \( \frac{dy}{dx} = y(\ln(x) + 1) \) Replacing y with the original function, we have: \( \frac{dy}{dx} = x^x(\ln(x) + 1) \) Now we have the first derivative, \(y' = x^x(\ln(x) + 1)\).
05

Differentiate the first derivative to find y''

Now we need to find the second derivative, \(y''\), by differentiating the first derivative with respect to x. For this, we can use the product rule as follows: \( \frac{d^2y}{dx^2} = \frac{d}{dx}(x^x(\ln(x) + 1)) \) \( \frac{d^2y}{dx^2} = (x^x \ln(x) + x^x) \frac{d}{dx}(\ln(x) + 1) + (\ln(x) + 1) \frac{d}{dx}(x^x \ln(x) + x^x) \) Now we need to find the derivatives involved in the equation above.
06

Calculate necessary derivatives

We need to find the derivative of two functions: 1. \(\frac{d}{dx}(\ln(x) + 1) = \frac{1}{x}\) 2. \(\frac{d}{dx}(x^x \ln(x) + x^x) = x^x(\ln(x) + 2)\), using the result from Step 4.
07

Substitute and simplify the second derivative expression

Now we substitute the above derivatives into the expression for the second derivative obtained in Step 5: \( \frac{d^2y}{dx^2} = (x^x \ln(x) + x^x) (\frac{1}{x}) + (\ln(x) + 1) (x^x(\ln(x) + 2) ) \) Simplifying the expression: \( \frac{d^2y}{dx^2} = x^{x-1} \ln(x) + x^{x-1} + x^x(\ln(x)+ 1)^2 \) And finally, we have found the second derivative of the given function: \( y'' = x^{x-1} \ln(x) + x^{x-1} + x^x(\ln(x) + 1)^2 \)

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Key Concepts

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

Second Derivative
Finding the second derivative is like peeking into the deeper behavior of functions. It tells us how the slope (first derivative) of a function is changing. In simpler terms, it helps us understand the curvature of the graph.
  • The process involves differentiating the first derivative. For example, if you take the first derivative, you get the rate of change. Taking the second derivative allows you to see how that rate itself is changing.
  • In our example, once we have the first derivative of the function \( y = x^x \), we apply differentiation again.
  • Why is this important? In mathematics and physics, understanding how things change over time or space is crucial. Second derivatives can indicate acceleration or concavity.
In calculus, recognizing the need for a second derivative often suggests that we're diving into more complex behavioral changes of functions, a valuable tool in many fields like engineering and economics.
Product Rule
The product rule helps when you have two functions multiplied together, and you want to find the derivative of their product. It's essential when dealing with functions that aren't simple products.To apply the product rule:1. Identify the two functions you're multiplying. Let's call them \( u \) and \( v \). 2. The derivative of their product is given by: \[ \frac{d}{dx}(uv) = u'v + uv' \]3. Apply this to both functions. Find the derivative of each function individually, then plug into the product rule formula.In our exercise:- We decomposed \( x^x(\ln(x) + 1) \) into two parts.- By utilizing the product rule, we accurately derived expressions for both the first and second derivatives.This rule simplifies the process of dealing with complex multiplications and ensures that we account for all changes when differentiating.
Natural Logarithm
A natural logarithm is a specific log with base \( e \), where \( e \) is approximately 2.718. It's used extensively in mathematics, especially calculus, due to its unique properties.
  • Natural logs simplify many expressions, especially when differentiating or integrating. For example, \( \ln(a^b) = b\ln(a) \) is a powerful property that helps in simplification.
  • The inverse relation with the exponential function \( e^x \) makes transformations easier, especially when dealing with continuous growth processes.
In the example, we used logarithmic differentiation to handle \( x^x \). Applying natural logs transformed the exponentiation into a multiplication, making differentiation straightforward.
Calculus
Calculus is a vast area of mathematics focused on change and motion. It allows us to calculate changes in quantities and understand the area under curves or the slopes of tangents. There are two fundamental branches:
  • Differential Calculus: Focuses on finding derivatives, which tell us the rate of change of a quantity.
  • Integral Calculus: Involves integrating functions, which helps in finding total quantities over an interval.
In our original exercise:- We used differential calculus to find both first and second derivatives of \( x^x \).- This use of calculus offers insights into the behavior of complex functions and helps solve real-world problems efficiently.With these tools, calculus not only assists in making predictions but also provides a deeper understanding of the natural and constructed world.

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Most popular questions from this chapter

Forecasting Commodity Crops Government economists in a certain country have determined that the demand equation for soybeans is given by $$ p=f(x)=\frac{55}{2 x^{2}+1} $$ where the unit price \(p\) is expressed in dollars per bushel and \(x\), the quantity demanded per year, is measured in billions of bushels. The economists are forecasting a harvest of \(2.2\) billion bushels for the year, with a possible error of \(10 \%\) in their forecast. Determine the corresponding error in the predicted price per bushel of soybeans.

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