Problem 50 - Calculus Early Transcendentals

Question

Compute the following derivatives using the method of your choice. \(\frac{d}{d x}\left(x^{2 x}\right)\)

Step-by-Step Solution

Answer

Answer: The derivative of the function \(x^{2x}\) is \(\frac{d}{dx}\left(x^{2x}\right) = x^{2x}(2\ln(x) + 2)\).

Step 1: Apply logarithmic differentiation

First, take the natural logarithm of both sides of the equation: $$\ln(y) = \ln\left(x^{2x}\right)$$ Where \(y = x^{2x}\).

Step 2: Simplify the equation

Use the power rule of logarithms to simplify the equation: $$\ln(y) = 2x\ln(x)$$

Step 3: Differentiate both sides with respect to x

Differentiate both sides of the equation with respect to x. On the left side, we have \(\frac{dy}{dx}\), and on the right side, we'll use the product rule: $$\frac{d}{dx}(\ln(y)) = \frac{d}{dx}(2x\ln(x))$$ We know that \(\frac{dy}{dt}\cdot\frac{dt}{dx}= \frac{dy}{dx}\) this allows us to rewrite $$\frac{1}{y}\cdot\frac{dy}{dx} = \frac{d}{dx}(2x\ln(x))$$ Now for the right-hand side, apply the product rule \((uv)' = u'v + uv'\): $$\frac{1}{y}\cdot\frac{dy}{dx} = 2\ln(x) + 2x\cdot\frac{1}{x}$$

Step 4: Solve for \(\frac{dy}{dx}\)

Now, we will solve for \(\frac{dy}{dx}\) by multiplying both sides with y: $$\frac{dy}{dx} = y\left(2\ln(x) + 2x\cdot\frac{1}{x}\right)$$

Step 5: Substitute back the original function

Recall that y is equal to \(x^{2x}\), so we will substitute this back into the equation to find the final derivative: $$\frac{dy}{dx} = x^{2x}\left(2\ln(x) + 2\right)$$ Therefore, the derivative of the given function is: $$\frac{d}{dx}\left(x^{2x}\right) = x^{2x}(2\ln(x) + 2)$$

Key Concepts

Logarithmic Differentiation

Logarithmic differentiation is a handy technique used especially when dealing with functions that have variables both in the base and the exponent, such as the expression \(x^{2x}\). By using logarithms, we can transform the expression into a more manageable form for differentiation. This involves taking the natural logarithm (ln) on both sides of the equation. Doing so allows us to use logarithmic properties to simplify the problem.

In this situation,

• We start by letting \(y = x^{2x}\).
• We then take the natural log of \(y\) to get \(\ln(y) = \ln(x^{2x})\).
• Thanks to the power rule of logarithms, which states that \(\ln(a^b) = b\ln(a)\), we simplify this to \(\ln(y) = 2x\ln(x)\).

This step breaks down the complexities caused by the variable in the exponent, making differentiation straightforward in the subsequent steps.

Power Rule

The power rule is one of the simplest and most widely used rules in differentiation. It allows us to find the derivative of a polynomial function with ease. If you have a function of the form \(f(x) = x^n\), where \(n\) is any real number, the power rule tells us that the derivative is \(f'(x) = nx^{n-1}\).

In the context of our problem, the power rule comes into play when simplifying the logarithmic expression into \(2x\ln(x)\). While using logarithmic differentiation, the exponent is moved in front of the log, thanks to the logarithm property, making it similar to a polynomial.

• Here, we don't directly use the power rule to differentiate \(x^{2x}\) as a whole.
• Instead, we prepare the function by creating a product of terms through logarithmic simplification.

This allows us to apply other differentiation rules more effectively.

Product Rule

The product rule is essential when taking the derivative of functions that are the product of two or more expressions. If you have two functions \(u(x)\) and \(v(x)\), the product rule states that the derivative of their product \(f(x) = u(x)v(x)\) is \[(u(x)v(x))' = u'(x)v(x) + u(x)v'(x)\]
In our exercise, after applying logarithmic differentiation and the power rule, the expression simplifies to \(2x\ln(x)\), a product of \(2x\) and \(\ln(x)\).

This is where the product rule comes in:

• Differentiate each part separately: The derivative of \(2x\) is \(2\), and the derivative of \(\ln(x)\) is \(1/x\).
• Apply the product rule: We get \(2\ln(x) + 2x\cdot 1/x = 2\ln(x) + 2\).

Thus, using the product rule allows us to handle the multiplication of different parts smoothly, moving us closer to the final derivative.

Chain Rule

The chain rule provides a method for differentiating composite functions, where one function is nested inside another. If you have a function \(y = f(g(x))\), the chain rule tells you to take the derivative of the outside function evaluated at the inside function times the derivative of the inside function. That is, \[\frac{dy}{dx} = f'(g(x))\cdot g'(x)\]

In the problem of finding \(\frac{d}{dx}(x^{2x})\), we use the chain rule implicitly by differentiating \(\ln(y)\) with respect to \(y\) and then multiplying by \(\frac{dy}{dx}\) due to the composite nature of our function expressions. This occurs after applying logarithmic differentiation, resulting in

• \(\frac{1}{y} \cdot \frac{dy}{dx}\).
• Essentially \(y\) is tackled as a composition of exponential and logarithmic functions, allowing clear differentiation.

Hence, the chain rule helps manage these nested forms during differentiation, forming a bridge from our transformed logarithmic expression to the final derivative formula.