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

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

Step 1: Apply Logarithmic Differentiation

(We will first apply the natural logarithm (ln) to both sides of the function equation and simplify it.) Assume a function: $$y = x^{2x}$$ Applying the natural logarithm (ln) to both sides: $$\ln{y} = \ln{x^{2x}}$$ By using the property of logarithms (the exponent in a log can be moved as a coefficient in front of the log), the expression becomes: $$\ln{y} = 2x \cdot \ln{x}$$

Step 2: Derive both sides with respect to x

(Now, we will differentiate both sides of the function equation with respect to x.) Derivative of the left side (using chain rule, where \(\frac{d}{dy}\) is the derivative of ln(y) with respect to y, and then we multiply by \(\frac{dy}{dx}\)): $$\frac{d}{d x}(\ln{y}) = \frac{1}{y}\cdot\frac{dy}{dx}$$ Derivative of the right side (using product rule): $$\frac{d}{d x}(2x \cdot \ln{x}) = 2 \cdot \ln{x} + 2x \cdot \frac{1}{x}$$

Step 3: Solve for the original derivative

(After differentiating both sides, we will solve for the derivative of the original function by isolating \(\frac{dy}{dx}\)) $$\frac{1}{y}\cdot\frac{dy}{dx} = 2 \cdot \ln{x} + 2x \cdot \frac{1}{x}$$ Multiplying both sides by y: $$\frac{dy}{dx} = y(2 \cdot \ln{x} + 2x \cdot \frac{1}{x})$$ Recall that \(y = x^{2x}\), replace y by the original function: $$\frac{dy}{dx} = x^{2x}(2 \cdot \ln{x} + 2)$$

Step 4: Final Answer

(The result is the derivative of the original function.) Now we have our final answer: $$\frac{d}{dx}\left(x^{2 x}\right) = x^{2x}(2 \cdot \ln{x} + 2)$$

Key Concepts

Logarithmic Differentiation

Logarithmic differentiation is a method used to simplify the process of finding derivatives, especially when you have complicated functions raised to the power of a variable. It can be particularly handy when dealing with functions like \(x^{2x}\).

Here is how it works:

• Start by taking the natural logarithm (ln) of both sides of your function: if \(y = x^{2x}\), then apply natural logarithm as \(\ln{y} = \ln{x^{2x}}\).
• Use the property of logarithms which states that the log of a power can be rewritten: in this case \(\ln{x^{2x}} = 2x \cdot \ln{x}\).
• Differentiate both sides with respect to \(x\) using implicit differentiation.

This approach transforms products and powers into more manageable sums and allows applying simpler calculus rules like the chain and product rule.
By working with the natural log, you make differentiation more straightforward, especially for products and quotients.

Chain Rule

The chain rule is an essential concept in calculus for computing the derivatives of composite functions. Whenever a function is composed of other functions, the chain rule can be applied.

Simply put, if you have a function \(y = f(g(x))\), the derivative of \(y\) with respect to \(x\) is obtained by multiplying the derivative of \(f\) with respect to \(g(x)\) by the derivative of \(g\) with respect to \(x\).

Here's the formula: \[\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\]This is particularly useful in our earlier step where we used implicit differentiation as part of logarithmic differentiation.

• Apply it to ln functions, such as \(\frac{d}{dy}(\ln{y}) = \frac{1}{y}\cdot\frac{dy}{dx}\); here, you're differentiating the inner \(y(x)\)

With the chain rule, handling nested functions and exponential functions becomes easier.

Product Rule

Another crucial calculus rule is the product rule, which is used when differentiating two multiplied functions. The product rule states: 'The derivative of a product is the derivative of the first times the second, plus the first times the derivative of the second.'

If \(u(x)\) and \(v(x)\) are two functions, then the derivative \( (uv)'(x) \) is: \[(uv)' = u'v + uv'\]

In our problem with \(2x \cdot \ln{x}\), the product rule helps break down the differentiation process like this:

• \(u(x) = 2x\), hence \(u'(x) = 2\)
• \(v(x) = \ln{x}\), hence \(v'(x) = \frac{1}{x}\)

Combine these to get: \[2 \cdot \ln{x} + 2x \cdot \frac{1}{x}\]

By using the product rule, we efficiently calculate derivatives when dealing with products of two functions, even if they're not as straightforward as basic polynomials.

Exponential Functions

Exponential functions occur when a variable's exponent drives the growth or decay rate across the function. They take the form \(f(x) = a^{x}\) but can also appear in forms combining base and variable exponents like \(x^{2x}\).

Here, \(x^{2x}\) is not a standard exponential function; however, logarithmic differentiation enables efficient handling of such cases. These functions showcase unique properties, like the natural exponential rate of change.

• To differentiate \( x^{2x} \), taking the natural logarithm eases the process.
• We transform to a more standard form and then revert back after finding the derivative.

Exponential functions often appear in real-world contexts such as compound interest, population growth, and radioactive decay. Their derivatives help describe how the function output changes upon each infinitesimally small input shift, crucial for modeling and predictions.