Problem 349 - Calculus Volume 1

Question

Use logarithmic differentiation to find \(\frac{d y}{d x}\). $$ y=x^{\log _{2} x} $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{2 x^{\log_{2} x - 1} \ln x}{\ln 2} \).

Step 1: Take the natural logarithm of both sides

Start by taking the natural logarithm of both sides of the equation to make differentiation easier: \( \ln(y) = \ln(x^{\log_{2} x}) \). Using the logarithm power rule, simplify the right-hand side: \( \ln(y) = \log_{2} x \cdot \ln x \).

Step 2: Express \( \log_{2} x \) in terms of natural logarithm

Rewrite \( \log_{2} x \) using the change of base formula: \( \log_{2} x = \frac{\ln x}{\ln 2} \). Substitute this into your equation for \( \ln(y) \): \( \ln(y) = \frac{\ln^2 x}{\ln 2} \).

Step 3: Differentiate both sides with respect to \( x \)

Use implicit differentiation on \( \ln(y) = \frac{\ln^2 x}{\ln 2} \). Differentiate the left side: \( \frac{d}{dx} \ln(y) = \frac{1}{y}\frac{dy}{dx} \). Now differentiate the right side: the derivative of \( \frac{\ln^2 x}{\ln 2} \) is \( \frac{2 \ln x}{\ln 2x} \cdot \frac{1}{x} \).

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

Set the derivatives equal: \[ \frac{1}{y} \frac{dy}{dx} = \frac{2 \ln x}{x \ln 2} \]. Multiply both sides by \( y \) to solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = y \cdot \frac{2 \ln x}{x \ln 2} \].

Step 5: Substitute \( y = x^{\log_{2} x} \) back into the equation

Replace \( y \) with the original function: \( \frac{dy}{dx} = x^{\log_{2} x} \cdot \frac{2 \ln x}{x \ln 2} \). Simplify the expression to get the final derivative: \( \frac{dy}{dx} = \frac{2 x^{\log_{2} x - 1} \ln x}{\ln 2} \).

Key Concepts

Derivative

A derivative is a fundamental concept in calculus that measures how a function changes as its input changes. In essence, it represents the rate of change or the slope of the function's graph at any given point. When we talk about finding derivatives, we often refer to the process of differentiation, which is the act of computing a derivative of a function.
Understanding derivatives is crucial as they help us solve various problems involving dynamics, optimization, and analysis of functions. When applying derivatives, remember a few key points:

• If a function describes a straight line, the derivative is constant and equal to the slope of the line.
• For non-linear functions, the derivative varies depending on the point on the curve.
• The derivative provides insight into the behavior of the function, including points where the function increases, decreases, or has critical points like maxima or minima.

With these concepts in mind, derivatives become a powerful tool for explaining changes in mathematical functions and real-world phenomena.

Change of Base Formula

The change of base formula is a handy tool for calculating logarithms in terms of different bases. Since many calculators and most mathematical analysis tools compute logarithms only for the base 10 or natural log (base \(e\)), converting other bases into these forms is necessary.
The formula is expressed as:\[\log_b a = \frac{\log_c a}{\log_c b}\]This formula allows you to convert a logarithm from one base \(b\) to another base \(c\), where you can choose \(c\) as 10 or \(e\) to facilitate computation.
Here’s a practical breakdown of using the change of base formula:

• Identify the logarithmic expression you need to convert, say \(\log_2 x\).
• Choose \(c\) as a base that is easier to calculate, typically base 10 or \(e\).
• Apply the formula: \(\log_2 x = \frac{\ln x}{\ln 2}\).

By changing the base, problems become more manageable especially when dealing with complex functions in calculus, such as logarithmic differentiation.

Implicit Differentiation

Implicit differentiation is a technique used to find derivatives of functions that are not explicitly solved for one variable in terms of another. This method is especially useful for dealing with equations where separating the variables is cumbersome or impossible. In implicit differentiation, you differentiate each term of the equation with respect to the variable of interest (often \(x\)) and apply the chain rule where needed.
Here’s how you can approach implicit differentiation:

• Begin by differentiating both sides of the equation with respect to \(x\).
• Wherever \(y\) appears, apply the chain rule: treat \(y\) as a function of \(x\) (\(y(x)\)).
• Every time you differentiate a term containing \(y\), multiply by \(\frac{dy}{dx}\) to account for the dependence of \(y\) on \(x\).
• Simplify the resulting expressions and solve for \(\frac{dy}{dx}\) to find the derivative.

Implicit differentiation is particularly powerful when used in combination with logarithmic differentiation strategies.

Power Rule

The power rule is a basic but important rule in calculus for finding the derivative of power functions. If you have a function \(f(x) = x^n\), where \(n\) is a real number, the power rule states that the derivative \(f'(x)\) is:\[f'(x) = n \, x^{n-1}\]This rule allows for quick differentiation of polynomial functions and terms where the variable \(x\) is raised to a power.
Here’s a step-by-step usage of the power rule:

• Identify any power term in the function, such as \(x^5\) or \(x^{rac{3}{2}}\).
• Apply the power rule by bringing down the exponent as a multiplier.
• Reduce the exponent by one to get the new power of \(x\).

The power rule simplifies differentiation, especially when paired with other rules like the product or quotient rule, enabling you to differentiate more complex functions easily.