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Chapter 3: Problem 86

Compute the following derivatives. Use logarithmic differentiation where appropriate. $$\frac{d}{d x}(2 x)^{2 x}$$

Short Answer

Expert verified
Answer: The derivative of the function \(y = (2x)^{2x}\) is \(\frac{dy}{dx} = (2x)^{2x}(2\ln(2x) + 2)\).

Step by step solution

01

Apply natural logarithm and simplify

Apply the natural logarithm (ln) function to both sides of the equation \(y = (2x)^{2x}\): $$ \ln(y) = \ln\left((2x)^{2x}\right) $$ Using logarithm properties, we can simplify the equation: $$ \ln(y) = 2x \ln(2x) $$
02

Differentiate both sides with respect to \(x\)

Differentiate both sides of the equation with respect to \(x\): $$ \frac{d}{dx} (\ln(y))= \frac{d}{dx} (2x \ln(2x)) $$ On the left side, apply the chain rule (\(\frac{d}{dx}(\ln(y))=\frac{1}{y}\cdot\frac{dy}{dx}\)): $$ \frac{1}{y}\cdot\frac{dy}{dx}= \frac{d}{dx} (2x \ln(2x)) $$
03

Differentiate the right side using the product rule

Differentiate \(2x \ln(2x)\) with respect to \(x\) using the product rule: $$ \frac{d}{dx} (2x \ln(2x)) = (2x)'\ln(2x) + 2x(\ln(2x))' $$ Differentiate \(2x\) and \(\ln(2x)\) separately: $$ (2x)' = 2 \quad \text{and} \quad (\ln(2x))' = \frac{1}{2x}\cdot2 = \frac{1}{x} $$ Substitute these values back into the product rule: $$ \frac{d}{dx} (2x \ln(2x)) = 2\ln(2x) + 2x\cdot\frac{1}{x} = 2\ln(2x) + 2 $$
04

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

We had previously found that: $$ \frac{1}{y}\cdot\frac{dy}{dx} = 2\ln(2x) + 2 $$ Solving for \(\frac{dy}{dx}\), we get: $$ \frac{dy}{dx} = y(2\ln(2x) + 2) $$
05

Replace \(y\) with the original function

Recall that \(y = (2x)^{2x}\). Substitute this back in: $$ \frac{dy}{dx} = (2x)^{2x}(2\ln(2x) + 2) $$ That's the final expression for the derivative.

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Key Concepts

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

Chain Rule
The chain rule is an essential technique in calculus used for differentiating composite functions. When you have a function inside another function, the chain rule allows you to find the derivative of the entire expression efficiently.
Understanding this concept is critical when dealing with functions where one variable depends on another, and in logarithmic differentiation, this rule plays a crucial role.Here's the basic idea:
  • Suppose we have a function \(y = f(g(x))\), where \(g(x)\) is another function inside \(f\).
  • The chain rule states that the derivative of \(y\) with respect to \(x\) is \(f'(g(x)) \cdot g'(x)\).
When applied to logarithmic differentiation, the chain rule becomes indispensable.
For example, examining the given problem, when differentiating the left side, we applied the chain rule to find \( \frac{d}{dx}(\ln(y))=\frac{1}{y}\cdot\frac{dy}{dx} \).
This application of the chain rule helps to simplify the complex nested functions often involved in these problems.
Product Rule
In calculus, the product rule is applied when you need to differentiate an expression involving the multiplication of two functions.
It's particularly useful for handling derivatives of expressions like \(uv\), where \(u\) and \(v\) are functions of \(x\).The product rule is expressed as:
  • \(\frac{d}{dx}(uv) = u'v + uv'\)
This rule ensures that both products of the derivatives are calculated and added together.
In the context of logarithmic differentiation and the original exercise, the problem requires us to differentiate \(2x \ln(2x)\), employing the product rule. First, we find the derivatives of each part:
  • The derivative of \(2x\) is \(2\).
  • The derivative of \(\ln(2x)\) is \(\frac{1}{x}\), achieved by applying the chain rule as well.
By substituting these derivatives into the product rule formula, we can correctly find the derivative of the entire expression, which simplifies the problem-solving process.
Natural Logarithm (ln)
The natural logarithm, denoted as \(\ln\), is a logarithm to the base \(e\), where \(e\) is an irrational constant approximately equal to 2.718.
The natural logarithm is fundamental in calculus due to its unique properties, particularly regarding derivatives and integration.
When dealing with logarithmic differentiation, \(\ln\) is extremely handy. This stems from a property that makes differentiation simpler:
  • The derivative of \(\ln(x)\) is \(\frac{1}{x}\).
This makes \(\ln\) preferable when solving problems involving exponentials or ratios.
In the original problem, by taking the natural logarithm of both sides of the equation \(y = (2x)^{2x}\), we could use \(\ln\) properties to transform the equation into a more manageable form:
  • Using \(\ln((2x)^{2x}) = 2x \ln(2x)\), simplifies the multiplication into an addition problem which is easier to derive.
Overall, understanding \(\ln\) and its application in calculus problems can significantly streamline finding derivatives, particularly when base-changing or dealing with intricate exponentials.

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Most popular questions from this chapter

Proof by induction: derivative of \(e^{k x}\) for positive integers \(k\) Proof by induction is a method in which one begins by showing that a statement, which involves positive integers, is true for a particular value (usually \(k=1\) ). In the second step, the statement is assumed to be true for \(k=n\), and the statement is proved for \(k=n+1,\) which concludes the proof. a. Show that \(\frac{d}{d x}\left(e^{k x}\right)=k e^{k x}\) for \(k=1\) b. Assume the rule is true for \(k=n\) (that is, assume \(\left.\frac{d}{d x}\left(e^{n x}\right)=n e^{n x}\right),\) and show this implies that the rule is true for \(k=n+1 .\) (Hint: Write \(e^{(n+1) x}\) as the product of two functions, and use the Product Rule.)

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. \(y=m x ; x^{2}+y^{2}=a^{2},\) where \(m\) and \(a\) are constants

Special Product Rule In general, the derivative of a product is not the product of the derivatives. Find nonconstant functions \(f\) and \(g\) such that the derivative of \(f g\) equals \(f^{\prime} g^{\prime}\).

Use any method to evaluate the derivative of the following functions. $$f(x)=\frac{4-x^{2}}{x-2}$$

Find \(f^{\prime}(x), f^{\prime \prime}(x),\) and \(f^{\prime \prime \prime}(x)\) \(f(x)=\frac{1}{x}\)
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