Problem 63 - Calculus for Biology and Medicine

Question

Use logarithmic differentiation to find the first derivative of the given functions. $$ f(x)=2 x^{x} $$

Step-by-Step Solution

Answer

The derivative of \( f(x) = 2x^x \) is \( 2x^x (\ln(x) + 1) \).

Step 1: Rewrite the Function Using Logarithms

To begin, take the natural logarithm of both sides of the equation, \( y = 2x^x \), giving us \( \ln(y) = \ln(2x^x) \). We can rewrite this as \( \ln(y) = \ln(2) + \ln(x^x) \). Applying the power rule for logarithms, \( \ln(x^x) = x \ln(x) \). Thus, we have \( \ln(y) = \ln(2) + x \ln(x) \).

Step 2: Differentiate Both Sides

Differentiate both sides of the equation with respect to \( x \). On the left, we use implicit differentiation to find \( \frac{d}{dx}[\ln(y)] = \frac{1}{y} \cdot \frac{dy}{dx} \). On the right, differentiate \( \ln(2) + x \ln(x) \) with respect to \( x \): \( \frac{d}{dx}[\ln(2)] = 0 \) and \( \frac{d}{dx}[x \ln(x)] = 1\cdot\ln(x) + x\cdot\frac{1}{x} = \ln(x) + 1 \) using the product rule. So, \( \frac{1}{y} \frac{dy}{dx} = \ln(x) + 1 \).

Step 3: Solve for the Derivative

Now, solve for \( \frac{dy}{dx} \) by multiplying both sides of the equation by \( y \): \( \frac{dy}{dx} = y(\ln(x) + 1) \). Substitute \( y = 2x^x \) back, giving \( \frac{dy}{dx} = 2x^x (\ln(x) + 1) \). Therefore, the derivative is \( \frac{dy}{dx} = 2x^x \ln(x) + 2x^x \). This is \( 2x^x (\ln(x) + 1) \) when factorized.

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique used when a function is not given in an explicit form, meaning it's not solved for one variable in terms of another. In cases where the relationship between variables is implicit, this method becomes very handy. Instead of solving for one variable and then differentiating, we differentiate directly.In our exercise, after taking the logarithm of both sides, we needed to differentiate with respect to \(x\). Because \(y\) is a function of \(x\), explicit differentation is not an option, thus implicit differentiation comes into play. When we differentiate \(\ln(y)\) with respect to \(x\), we use the chain rule: the derivative of \(\ln(y)\) is \(\frac{1}{y}\), and then multiply this by \(\frac{dy}{dx}\) because of the chain rule. By allowing this method, we can handle complex relationships between variables naturally.

Natural Logarithm

A natural logarithm is a logarithm with base \(e\), where \(e\) is an irrational and transcendental number approximately equal to 2.71828. It's an important concept in calculus, especially when it comes to manipulating exponentials. The natural logarithm has useful properties that simplify calculations, such as the power rule for logarithms.In our problem, we applied the natural logarithm to the function \(y = 2x^x\), transforming it into \(\ln(y) = \ln(2) + x\ln(x)\). This powerful step translates the problem into a sum of simpler parts, making differentiation feasible with rules like the constant rule and the product rule. This ability to break down complex expressions into manageable pieces is one of the key strengths of using natural logarithms in calculus.

Product Rule

The product rule is an essential technique in calculus used when differentiating products of functions. When you have a function \(u(x)\cdot v(x)\), the derivative is given by \(u'(x)v(x) + u(x)v'(x)\). This rule helps in handling functions that cannot easily be separated into simpler terms.In the solution, one part to differentiate is \(x \ln(x)\). By observing this as a product where \(u(x) = x\) and \(v(x) = \ln(x)\), we apply the product rule as follows:

• The derivative of \(x\) with respect to \(x\) is 1.
• The derivative of \(\ln(x)\) with respect to \(x\) is \(\frac{1}{x}\).

Plugging these into the product rule formula, we find that:\[ u'(x)v(x) + u(x)v'(x) = 1\cdot\ln(x) + x\cdot\frac{1}{x} = \ln(x) + 1 \]This simplification made through the product rule helps to capture the differentials' changes needed for further steps.

Derivative Calculation

Calculating the derivative using the tools we've discussed enables us to find the rate of change of the function at any given point. Let's summarize the steps taken once the transformations and differentiation rules were applied: - The implicit differentiation gave us the left side as \(\frac{1}{y}\frac{dy}{dx}\). - The product rule and natural logarithm simplifications determined the right side to be \(\ln(x) + 1\).To isolate \(\frac{dy}{dx}\), we multiply both sides by \(y\):\[ \frac{dy}{dx} = y(\ln(x) + 1) \]Substituting back \(y = 2x^x\), we complete the derivative calculation:\[ \frac{dy}{dx} = 2x^x (\ln(x) + 1) \]This gives the first derivative of the function \(f(x)=2x^x\). This result not only provides the slope of the tangent line to the curve at any point \(x\), but makes use of intricate rules like implicit differentiation and the product rule, showcasing the interconnected nature of calculus.