Question

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

Short answer

\( \frac{dy}{dx} = x^{\log_2 x} \left(\frac{\ln x}{x \ln 2} + \frac{\log_2 x}{x}\right) \)

Step-by-step solution

Apply Natural Logarithm to Both Sides

Take the natural logarithm of both sides of the equation to simplify the differentiation process. Start with:\[ \ln y = \ln (x^{\log_2 x}) \]Use the power rule for logarithms, \(\ln(a^b) = b \cdot \ln a\), to rewrite the right side:\[ \ln y = (\log_2 x) \cdot \ln x \]

Differentiate Both Sides Using Implicit Differentiation

Differentiate both sides of the equation with respect to \(x\). For the left side, use the derivative of \(\ln y\), which is \(\frac{1}{y} \cdot \frac{dy}{dx}\). For the right side, apply the product rule. \(\frac{d}{dx}(\ln y) = \frac{d}{dx} ((\log_2 x) \cdot \ln x)\)The differentiation of the right side using the product rule and logarithm derivatives yields:\[ \frac{1}{y} \cdot \frac{dy}{dx} = \frac{d}{dx}(\log_2 x) \cdot \ln x + \log_2 x \cdot \frac{d}{dx}(\ln x) \]

Calculate the Derivatives on Right Side

Calculate each derivative separately:1. \(\frac{d}{dx}(\log_2 x) = \frac{1}{x \ln 2}\) because the derivative of \(\log_b x\) is \(\frac{1}{x \ln b}\).2. \(\frac{d}{dx}(\ln x) = \frac{1}{x}\).Putting them together:\[ \frac{1}{y} \cdot \frac{dy}{dx} = \left(\frac{1}{x \ln 2}\right) \cdot \ln x + \left(\log_2 x\right) \cdot \frac{1}{x} \]Simplify:\[ \frac{1}{y} \cdot \frac{dy}{dx} = \frac{\ln x}{x \ln 2} + \frac{\log_2 x}{x} \]

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

Multiply both sides by \(y\) to isolate \(\frac{dy}{dx}\):\[ \frac{dy}{dx} = y \left(\frac{\ln x}{x \ln 2} + \frac{\log_2 x}{x}\right) \]Substitute back \(y = x^{\log_2 x}\) into the expression:\[ \frac{dy}{dx} = x^{\log_2 x} \left(\frac{\ln x}{x \ln 2} + \frac{\log_2 x}{x}\right) \]Hence, the derivative is expressed as given.

Key concepts

Implicit Differentiation

Implicit differentiation is a powerful tool used when dealing with equations where one variable is not explicitly solved in terms of the other. In our exercise, after taking the natural logarithm of both sides, we get an equation involving both \(y\) and its derivative. This is where implicit differentiation comes in.
To differentiate both sides of \(abla y = ( ext{log}_2 x) \cdot abla x\), we treat \(abla y\) as an implicit function of \(x\). This means we apply the derivative to \(y\) with respect to \(x\), acknowledging that \(y\) is a function of \(x\) itself. The left side differentiates to \(\frac{1}{y} \cdot \frac{dy}{dx}\).
By treating \(y\) implicitly, this differentiation gets us closer to expressing \(\frac{dy}{dx}\) in terms of \(x\), which is our ultimate goal in finding the derivative of the equation.

Power Rule for Logarithms

The power rule for logarithms is a simple yet essential property that greatly simplifies differentiation, especially in logarithmic differentiation scenarios. When we have an expression like \(a^b\), the power rule states that \(\ln(a^b) = b \cdot \ln a\).
This rule is precisely what allows us to handle the original equation in our problem. When we take the natural logarithm of both sides, \(\ln(x^{\log_2 x})\), the power rule transforms it into a more manageable form: \( (\log_2 x) \cdot \ln x\).
This transformation simplifies the differentiation process significantly, because now the expression on the right side is a product of two functions, \(\log_2 x\) and \(\ln x\), which we can tackle using the product rule for differentiation.

Product Rule

The product rule is a fundamental differentiation technique used when differentiating the product of two functions. If you have two functions \(u\) and \(v\), their derivative \(\frac{d}{dx}(uv)\) is given by the expression \(u'v + uv'\).
In our exercise, when we reach the step \(\frac{d}{dx} ((\log_2 x) \cdot \ln x)\), the product rule is crucial. With \(u = \log_2 x\) and \(v = \ln x\), we differentiate to find \(u'\) and \(v'\), which are \(\frac{1}{x \ln 2}\) and \(\frac{1}{x}\), respectively.
This results in \(\frac{1}{y} \cdot \frac{dy}{dx} = u'v + uv'\), which simplifies to \(\frac{\ln x}{x \ln 2} + \frac{\log_2 x}{x}\). By using the product rule, we can clearly and methodically differentiate and eventually solve for the derivative \(\frac{dy}{dx}\).

• Remember, the product rule only applies when dealing with multiplication of functions.
• Each term of the product needs to be differentiated separately, holding one function constant at a time.