Problem 350 - Calculus Volume 1

Question

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

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = (x^2 - 1)^{\ln x} \left( \ln(x^2 - 1) \cdot \frac{1}{x} + \ln x \cdot \frac{2x}{x^2 - 1} \right) \)

Step 1: Introduce Logarithmic Differentiation

Since the function involves a power of a function, logarithmic differentiation is useful. We take the natural logarithm on both sides: \[ \ln y = \ln \left((x^2 - 1)^{\ln x}\right) \]

Step 2: Simplify the Logarithm

Use the property of logarithms, \( \ln(a^b) = b \cdot \ln a \), to simplify the expression:\[ \ln y = \ln x \cdot \ln(x^2 - 1) \]

Step 3: Differentiate Both Sides

Differentiate both sides with respect to \( x \). Use the product rule on the right side:\[ \frac{d}{dx}(\ln y) = \frac{d}{dx} (\ln x \cdot \ln(x^2 - 1)) \]This gives:\[ \frac{1}{y} \cdot \frac{dy}{dx} = \ln(x^2 - 1) \cdot \frac{1}{x} + \ln x \cdot \frac{d}{dx} (\ln(x^2 - 1)) \]

Step 4: Differentiate the Remaining Logarithm

Compute the derivative of \( \ln(x^2 - 1) \) using the chain rule:\[ \frac{d}{dx} (\ln(x^2 - 1)) = \frac{1}{x^2 - 1} \cdot 2x \]

Step 5: Substitute Back into the Equation

Substitute the derivative from Step 4 back into the equation from Step 3:\[ \frac{1}{y} \cdot \frac{dy}{dx} = \ln(x^2 - 1) \cdot \frac{1}{x} + \ln x \cdot \frac{2x}{x^2 - 1} \]

Step 6: Solve for \( \frac{dy}{dx} \)

Multiply through by \( y \) to solve for \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = y \left( \ln(x^2 - 1) \cdot \frac{1}{x} + \ln x \cdot \frac{2x}{x^2 - 1} \right) \]Substitute back \( y = (x^2 - 1)^{\ln x} \):\[ \frac{dy}{dx} = (x^2 - 1)^{\ln x} \left( \ln(x^2 - 1) \cdot \frac{1}{x} + \ln x \cdot \frac{2x}{x^2 - 1} \right) \]

Key Concepts

Natural Logarithm

The natural logarithm, represented as \(\ln\), is the logarithm to the base of the mathematical constant \(e\), where \(e\) is approximately equal to 2.71828. The natural logarithm has several useful properties that make it an invaluable tool in calculus, particularly for simplifying expressions involved in differentiation and integration.
A powerful property of the natural logarithm is that the logarithm of a power can be simplified. For example, \( \ln(a^b) = b \cdot \ln a \). This property is exactly what we used in the exercise to simplify the expression \( \ln((x^2 - 1)^{\ln x}) \) to \( \ln x \cdot \ln(x^2 - 1) \).
By taking the natural logarithm of a function, we can turn products into sums, quotients into differences, and powers into products, making it easier to differentiate complex combinations of functions.

Product Rule

The product rule is an essential differentiation rule used when differentiating a product of two functions. It states that if you have two functions, \( u(x) \) and \( v(x) \), then the derivative of their product \( u(x)v(x) \) is given by:

• \( (uv)' = u'v + uv' \)

In our exercise, we had the product of two logs, \( \ln x \) and \( \ln(x^2 - 1) \). To differentiate this product, we applied the product rule.
The differentiation was done by first differentiating \( \ln x \), which gives \( \frac{1}{x} \), and then differentiating \( \ln(x^2 - 1) \), followed by combining these results according to the product rule.

Chain Rule

The chain rule is another fundamental rule used when the derivative of a composite function is needed. A composite function is a function inside another function, which can be expressed as \( g(f(x)) \). The chain rule states that to differentiate it, you perform:

• \( \frac{d}{dx}[g(f(x))] = g'(f(x)) \cdot f'(x) \)

In the exercise, we needed to differentiate \( \ln(x^2 - 1) \). Here, \( f(x) = x^2 - 1 \) and \( g(x) = \ln(x) \). Differentiating \( \ln(f(x)) \) involved finding \( \frac{1}{f(x)} \) and then multiplying by the derivative of \( f(x) \), which is \( 2x \).
Thus, the derivative of \( \ln(x^2 - 1) \) becomes \( \frac{2x}{x^2 - 1} \). Using the chain rule allows us to correctly obtain the derivative of a nested function.

Derivative Calculation

Calculating derivatives is a fundamental process in calculus, involving the computation of the instantaneous rate of change of a function. The exercise made use of logarithmic differentiation, a technique that uses the properties of logarithms and the rules of differentiation to make the calculation more manageable.
Initially, by taking the natural logarithm of the entire function and simplifying using logarithmic properties, we then differentiated both sides of the equation. This involved applying the product rule and the chain rule. Ultimately, solving for \( \frac{dy}{dx} \) involved isolating \( \frac{dy}{dx} \) on one side, and substituting back the original value of \( y \).
The final expression represented the derivative, showing how carefully applying these techniques leads to a successful differentiation of complex functions that would otherwise be challenging to handle directly.