Question

Use logarithmic differentiation to find \(\frac{d y}{d x}\). $$ y=\left(x^{2}-1\right)^{\ln x} $$

Short answer

The derivative is \((x^2-1)^{\ln x} \left( \frac{\ln(x^2-1)}{x} + \frac{2x \ln x}{x^2-1} \right)\).

Step-by-step solution

Introduce the Problem

We need to differentiate the function \(y = (x^2 - 1)^{\ln x}\) using logarithmic differentiation. Logarithmic differentiation is useful when the function is a power of a variable expression.

Take the Natural Logarithm

First, take the natural logarithm of both sides. This step will help simplify the differentiation process. We have: \[ \ln y = \ln((x^2-1)^{\ln x}) \] Using the property of logarithms that \(\ln(a^b) = b \ln a\), we simplify this to: \[ \ln y = \ln x \cdot \ln(x^2 - 1) \]

Differentiate Both Sides

Differentiate both sides of the equation with respect to \(x\). For the left side, use implicit differentiation: \[ \frac{1}{y} \frac{dy}{dx} \] For the right side, apply the product rule to differentiate \(\ln x \cdot \ln(x^2-1)\): \[ \frac{d}{dx} [\ln x \cdot \ln(x^2 - 1)] = \ln(x^2-1) \cdot \frac{1}{x} + \ln x \cdot \frac{2x}{x^2-1} \]

Simplify the Derivative

Simplify the expression obtained from Step 3: \[ \frac{1}{y} \frac{dy}{dx} = \frac{\ln(x^2 - 1)}{x} + \frac{2x \ln x}{x^2-1} \] Multiply through by \(y\) to solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = y \left( \frac{\ln(x^2 - 1)}{x} + \frac{2x \ln x}{x^2-1} \right) \]

Substitute Back for y

Remember that \(y = (x^{2} - 1)^{\ln x}\), substitute this back into the expression obtained in Step 4: \[ \frac{dy}{dx} = (x^2 - 1)^{\ln x} \left( \frac{\ln(x^2 - 1)}{x} + \frac{2x \ln x}{x^2-1} \right) \]

Key concepts

Implicit Differentiation

Implicit differentiation is a technique used when it's difficult or even impossible to solve an equation explicitly for one variable in terms of another. In many complex functions, especially those involving products or powers like the one in our exercise, implicit differentiation is handy. The main idea is that every function can be thought of implicitly as a function of another variable.For example, if we have an equation involving both \(y\) and \(x\), instead of solving for \(y\) in terms of \(x\), we can differentiate both sides with respect to \(x\). Here's how you can visualize it:

• Differentiate both sides of the equation with respect to \(x\) as usual.
• Whenever you differentiate something involving \(y\), include \(\frac{dy}{dx}\) multiplied by the derivative of the function with respect to \(y\).

Implicit differentiation allows us to handle functions that are not easy to rearrange, opening up a range of possibilities for finding derivatives.

Product Rule in Calculus

The product rule is a fundamental tool in calculus used to find the derivative of a product of two functions. If you have two differentiable functions, say \(u(x)\) and \(v(x)\), the product rule states that the derivative of the product \(u(x) \cdot v(x)\) is given as:\[ (u \cdot v)' = u' \cdot v + u \cdot v' \]In simple terms, you take the derivative of the first function, multiply it by the second function, and then add the product of the first function and the derivative of the second function.This rule is particularly useful when dealing with expressions that involve multiplying functions together, as in our exercise where we needed to differentiate \(\ln x \cdot \ln(x^2 - 1)\). Applying the product rule ensured that we correctly accounted for each part of the expression.

Derivative of Logarithmic Functions

Understanding how to differentiate logarithmic functions is essential, especially when applying logarithmic differentiation. The derivative of the natural logarithm function, \(\ln x\), is straightforward: it's simply \(\frac{1}{x}\). This result is critical when simplifying derivatives of more complicated functions, where you apply the chain rule and/or product rule together with logarithmic differentiation. By transforming power expressions into a product via logarithmic properties, you can then easily apply these differentiation rules.In our given problem, using the identity \(\ln((x^2-1)^{\ln x}) = \ln x \cdot \ln(x^2 - 1)\) allowed us to straightforwardly find derivatives, converting potentially tricky power expressions into more manageable terms.