/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 Evaluate the following derivativ... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate the following derivatives. $$\frac{d}{d x}\left(x^{2 x}\right)$$

Short Answer

Expert verified
Answer: The derivative of \(x^{2x}\) with respect to x is \(\frac{d}{dx}\left(x^{2x}\right) = x^{2x}(2\ln(x) + 2)\).

Step by step solution

01

Rewrite the function using natural logarithm

Begin by rewriting the given function, \(x^{2x}\), using the natural logarithm: Let \(y = x^{2x}\). Then, take the natural logarithm of both sides: $$\ln(y) = 2x \ln(x)$$
02

Differentiate both sides with respect to \(x\)

Now, we differentiate both sides of the equation with respect to \(x\), remembering to use the chain rule and product rule where necessary: $$\frac{1}{y} \frac{dy}{dx} = 2\ln(x) + 2x\frac{1}{x}$$ Here, we differentiated the left side, applying the chain rule as \(\frac{d \ln(y)} {dy} \cdot \frac{dy}{dx}\) and the right side using the product rule.
03

Solve for \(\frac{dy}{dx}\)

Next, we will isolate the term \(\frac{dy}{dx}\): $$\frac{dy}{dx} = y \left(2\ln(x) + \frac{2x}{x}\right)$$
04

Substitute \(y\) back into the equation

Since we know \(y = x^{2x}\), we can substitute this back into the equation to find the derivative in the original form: $$\frac{dy}{dx} = x^{2x} \left(2\ln(x) + 2\right)$$
05

Simplify the expression

Now, let's simplify the expression: $$\frac{d}{dx}\left(x^{2x}\right) = x^{2x}(2\ln(x) + 2)$$

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Key Concepts

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

Natural Logarithm
The natural logarithm, denoted as \(\ln(x)\), is the inverse operation of exponentiation with the number \(e\) as the base, where \(e\) is an irrational constant approximately equal to 2.71828. This special logarithm is significant in calculus because it has properties that make working with exponential functions manageable.

For instance, the derivative of the natural logarithm of \(x\), \(\frac{d}{dx}\ln(x)\), equals \(\frac{1}{x}\). When dealing with compound functions involving exponents, as shown in the example exercise, we often use the natural logarithm to express the function in a simpler form that's more conducive to differentiation. The relationship between the natural logarithm and exponential functions hence becomes a powerful tool in calculus for solving complex problems.

Key Properties

  • \(\ln(e) = 1\) because because the logarithm of the base to itself is always one.
  • \(\ln(1) = 0\) because any number to the power of zero equals one, including \(e\).
  • The natural logarithm turns multiplication into addition and division into subtraction, making it easier to differentiate products or quotients.
Chain Rule
The chain rule is a differentiation technique used to compute the derivative of a composite function. When one function is nested within another, the chain rule allows us to differentiate the entire function by taking the derivative of the outer function and multiplying it by the derivative of the inner function.

For the given exercise, we see the chain rule in action when differentiating \(\ln(y)\) with respect to \(x\). This is because \(y\) itself is a function of \(x\) (\(y = x^{2x}\)). We first differentiate \(\ln(y)\) with respect to \(y\) to obtain \(\frac{1}{y}\), and then we multiply that by \(\frac{dy}{dx}\), which is the derivative of \(y\) with respect to \(x\). The chain rule is vital to calculus as it bridges the gap between complex compound functions and their rates of change.

Remembering the Chain Rule

  • Look for functions within functions (composition).
  • Differentiate the outer function, keeping the inner function as is.
  • Multiply by the derivative of the inner function.
Product Rule
The product rule is a principle that guides us on how to take the derivative of a product of two functions. It states that the derivative of a product is the derivative of the first function multiplied by the second function, plus the first function times the derivative of the second function. Mathematically, for two functions \(u(x)\) and \(v(x)\), the product rule is given by \(\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)\).

In the exercise given, we apply the product rule to differentiate \(2x\ln(x)\). We consider \(2x\) and \(\ln(x)\) as the two functions and follow through with the rule to obtain \(2\ln(x) + 2\). Understanding and correctly applying the product rule is essential in calculus when dealing with functions that are presented as products of two or more terms.

Application Tips

  • Identify the two distinct functions being multiplied.
  • Apply the product rule formula carefully, noting each function's derivative.
  • Simplify the result whenever possible.
Differentiation Techniques
In calculus, differentiation techniques, including the chain rule and product rule, are strategies used to find the rate of change or slope of a curve at any point. These techniques are crucial for dealing with a variety of functions, whether they're simple polynomials, implicit functions, trigonometric, logarithmic, or exponential functions.

The exercise demonstrations how a combination of differentiation techniques can be applied to a complex function. We start with a logarithmic differentiation, which is not a rule but a technique that utilizes the natural logarithm to transform the function into a more manageable form. Then, we apply established rules like the chain rule and the product rule to differentiate the transformed equation. These techniques provide a structured approach to find derivatives, which are foundational to understanding motion, growth, and a multitude of other scenarios described by mathematical models.

Strategy Overview

  • Use logarithmic differentiation when dealing with powers that are functions.
  • Identify which rules apply to the function you are differentiating.
  • Combine rules effectively to simplify and differentiate complex functions.
  • Remember to simplify the final result and match it to the original function (if necessary).

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