Problem 33 - Calculus - AP Edition

Question

Find \(d y / d x\) using the method of logarithmic differentiation. $$y=(\ln x)^{\tan x}$$

Step-by-Step Solution

Answer

The derivative is \(\frac{dy}{dx} = (\ln x)^{\tan x} \left(\sec^2 x \cdot \ln(\ln x) + \frac{\tan x}{x \ln x}\right)\).

Step 1: Apply the logarithm on both sides

Take the natural logarithm of both sides to simplify the expression, \[ \ln y = \ln((\ln x)^{\tan x}) \]Using the power rule for logarithms, this becomes:\[ \ln y = \tan x \cdot \ln(\ln x) \].

Step 2: Differentiate implicitly

Differentiate both sides with respect to \(x\). The left side, \(\ln y\), becomes \(\frac{1}{y} \cdot \frac{dy}{dx}\). The right side requires the product rule for differentiation:\[ \frac{d}{dx}[\tan x \cdot \ln(\ln x)] = \frac{d}{dx}(\tan x) \cdot \ln(\ln x) + \tan x \cdot \frac{d}{dx}(\ln(\ln x)) \].

Step 3: Calculate derivatives

Calculate the derivatives:- \(\frac{d}{dx}(\tan x) = \sec^2 x\).- \(\frac{d}{dx}(\ln(\ln x)) = \frac{1}{\ln x} \cdot \frac{1}{x} = \frac{1}{x \ln x}\).Substitute these results into the expression from Step 2:\[ \frac{d}{dx}[\tan x \cdot \ln(\ln x)] = \sec^2 x \cdot \ln(\ln x) + \tan x \cdot \frac{1}{x \ln x} \].

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

We have:\[ \frac{1}{y} \cdot \frac{dy}{dx} = \sec^2 x \cdot \ln(\ln x) + \frac{\tan x}{x \ln x} \]Multiply both sides by \(y\) to solve for \(\frac{dy}{dx}\):\[ \frac{dy}{dx} = y \left(\sec^2 x \cdot \ln(\ln x) + \frac{\tan x}{x \ln x}\right) \].

Step 5: Substitute \(y\) back

Recall that \(y = (\ln x)^{\tan x}\). Substitute back into the equation from Step 4 to express the derivative entirely in terms of \(x\):\[ \frac{dy}{dx} = (\ln x)^{\tan x} \left(\sec^2 x \cdot \ln(\ln x) + \frac{\tan x}{x \ln x}\right) \].

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique used when you have an equation with dependent and independent variables that are not easily separated. Rather than solving for one variable in terms of the other, you differentiate both sides of the equation with respect to the independent variable directly. This approach helps to find derivatives of functions that are defined implicitly.

With implicit differentiation, the chain rule is crucial. When you differentiate a function of the dependent variable, treat this variable as a function of the independent variable and multiply by its derivative. For instance, in the exercise provided, \(y = (\ln x)^{\tan x}\) was transformed into an implicit form:

• The natural logarithm was applied to both sides.
• It changed the problem into \(\ln y = \tan x \cdot \ln(\ln x)\).

To differentiate \(\ln y\) with respect to \(x\), use \(\frac{1}{y} \cdot \frac{dy}{dx}\) because of the chain rule. This manipulation allows us to directly work with more complex functions, useful when algebraic manipulation alone becomes cumbersome.

This method is especially helpful for equations where conventional separation of variables isn't straightforward. It enhances the flexibility of derivative computation in complex functional relationships.

Product Rule

The Product Rule is a fundamental technique in calculus used to find the derivative of the product of two functions. If you have two differentiable functions, \(u(x)\) and \(v(x)\), the derivative of their product \(u(x)v(x)\) is given by:
\[ u'(x)v(x) + u(x)v'(x) \]This rule allows you to break down derivatives involving products into simpler components.

In the original problem, after taking the logarithm, we have a product \(\tan x \cdot \ln(\ln x)\). Here, the functions are \(u(x) = \tan x\) and \(v(x) = \ln(\ln x)\). Applying the product rule means differentiating \(\tan x\) and \(\ln(\ln x)\) independently, then combining these results as follows:

• Differentiate \(\tan x\) to get \(\sec^2 x\).
• Differentiate \(\ln(\ln x)\) to get \(\frac{1}{x \ln x}\).
• Combine using the product rule: \(\sec^2 x \cdot \ln(\ln x) + \tan x \cdot \frac{1}{x \ln x}\).

This technique makes it easier to handle complex differentiations where functions multiply, providing a systematic way to manage each function's derivative contribution.

Power Rule

The Power Rule is one of the simplest and most commonly used differentiation techniques. It states that if you have a function \(x^n\), where \(n\) is any real number, the derivative is \(nx^{n-1}\). This rule simplifies the process of finding the rate of change of polynomial expressions.

While the Power Rule typically deals with simple powers of \(x\), in logarithmic differentiation, we leverage a version of this rule. For an expression like \((\ln x)^{\tan x}\), the Power Rule is used in the context of logarithms. By first taking the natural log, the expression becomes a multiplication problem \(\tan x \cdot \ln(\ln x)\) and is then differentiated.

• The key insight here is using logarithmic identities and differentiation in tandem.
• It converts an expression that looks like a power into a format amenable to the application of other rules.

The Power Rule serves as a foundational step in this process, underpinning more complex differentiation techniques such as those used in the given exercise, making it a versatile tool in calculus.