Question

, find dy/dx by logarithmic differentiation. Find and simplify \(f^{\prime}(1)\) if $$ f(x)=\ln \left(\frac{a x-b}{a x+b}\right)^{c}, \text { where } c=\frac{a^{2}-b^{2}}{2 a b} . $$

Short answer

The simplified value of \( f'(1) \) is 1.

Step-by-step solution

Express the Function in Terms of Its Logarithm

The function we have is given by \( f(x) = \ln \left(\frac{a x - b}{a x + b}\right)^c \). By using logarithmic properties, we can rewrite this as \( f(x) = c \cdot \ln \left(\frac{a x - b}{a x + b}\right) \).

Differentiate Using the Chain Rule

To differentiate \( f(x) = c \cdot \ln \left(\frac{a x - b}{a x + b}\right) \), we apply the chain rule and the quotient rule. For the inner function \( \ln \left(\frac{a x - b}{a x + b}\right) \), differentiation yields:\[ \frac{d}{dx} \ln \left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{u v} \]where \( u = a x - b \) and \( v = a x + b \). Thus, \( u' = a \) and \( v' = a \).

Calculate the Derivative

Apply the derivative formula:\[\frac{d}{dx} \ln \left(\frac{a x - b}{a x + b}\right) = \frac{(a x + b) \cdot a - (a x - b) \cdot a}{(a x - b)(a x + b)} = \frac{ab + ab}{(a x - b)(a x + b)} = \frac{2ab}{(a x - b)(a x + b)}\]Thus, the derivative is:\[f'(x) = c \cdot \frac{2ab}{(a x - b)(a x + b)}\]Substitute \(c = \frac{a^2 - b^2}{2ab}\) to get:\[f'(x) = \left(\frac{a^2 - b^2}{2ab}\right) \cdot \frac{2ab}{(a x - b)(a x + b)} = \frac{a^2 - b^2}{(a x - b)(a x + b)}\]

Calculate \(f'(1)\)

Substitute \(x = 1\) into the expression for \(f'(x)\):\[ f'(1) = \frac{a^2 - b^2}{(a(1) - b)(a(1) + b)} = \frac{a^2 - b^2}{(a - b)(a + b)} \]This simplifies directly to \( \frac{a^2 - b^2}{a^2 - b^2} = 1 \).

Conclusion of \(f'(1)\)

By evaluating the expression at \(x = 1\), we find that \(f'(1) = 1\). This is the final simplified derivative at the point \(x = 1\).

Key concepts

Chain Rule

The chain rule is a fundamental tool in calculus used for differentiating composite functions. It helps you find the derivative when a function is nested inside another function. Put simply, if you have a function like \( f(g(x)) \), where \( f \) and \( g \) are both differentiable, the chain rule states: \[ (f(g(x)))' = f'(g(x)) \cdot g'(x). \]In our example, the function \( f(x) = c \cdot \ln\left(\frac{a x - b}{a x + b}\right) \) forms a composite function with the logarithm being the outer function and the fraction its inner function. Using the chain rule, we first differentiate the outer function with respect to its inner function and then multiply it by the derivative of the inner function. This process ensures we accurately capture the effect of both functions' behavior as \( x \) changes.

Quotient Rule

When you're dealing with functions that include quotients (fractions where the numerator and denominator are functions themselves), the quotient rule comes into play. The rule provides a way to find the derivative of a quotient \( \frac{u}{v} \). The quotient rule is given by:\[ \left( \frac{u}{v} \right)' = \frac{v \cdot u' - u \cdot v'}{v^2}. \]This rule was applied in differentiating the inner function of our composite function \( \ln\left(\frac{a x - b}{a x + b}\right) \). Here, \( u = a x - b \) and \( v = a x + b \) with their derivatives \( u' = a \) and \( v' = a \). By substituting these into the quotient rule formula, you ultimately reduce the complexity of the fraction, laying the groundwork for simpler further calculations. This step is pivotal for proceeding towards full derivative simplification.

Derivative Simplification

Once the chain and quotient rules are applied, derivative simplification follows. This involves combining and rearranging terms to attain a more manageable form of the derivative. In our exercise, after applying both the chain and quotient rules, the expression\[ \frac{d}{dx} \ln \left( \frac{a x - b}{a x + b} \right) = \frac{(a x + b) \cdot a - (a x - b) \cdot a}{(a x - b)(a x + b)} \] is simplified to:\[ \frac{2ab}{(a x - b)(a x + b)}. \]Simplification makes the expression cleaner and more efficient for further evaluations. In this case, we further substitute for \( c \) and simplify, optimizing the derivative for calculating specific values like \( f'(1) \). By simplifying we transform complex outputs into more practical terms, aiding both calculation efficiency and result interpretation.

Logarithm Properties

Logarithmic properties are various rules that help simplify expressions involving logs, making calculations easier. Key properties include the power rule which states that \( \ln(a^b) = b \cdot \ln(a) \) and the quotient rule that defines \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b). \)In the given exercise, this power property is used to express the original function as \( f(x) = c \cdot \ln\left(\frac{a x - b}{a x + b}\right) \). Using logarithmic properties effectively simplifies early steps in differentiation, allowing the function’s behavior to be more predictively managed. Understanding and applying these properties is crucial, especially in exercises involving logarithmic differentiation, as they streamline both the derivatives' formation and simplification process. Ultimately, these properties both facilitate easier differentiation and improve the clarity and manageability of final solutions.