Question

If \[ y = \left(\dfrac{x+1}{x-1}\right)^x, \] then find \(\dfrac{dy}{dx}\).

• A. \(y\left[\ln\left(\dfrac{x+1}{x-1}\right)-\dfrac{2x}{x^2-1}\right]\)
• B. \(y\left[\ln\left(\dfrac{x+1}{x-1}\right)+\dfrac{2x}{x^2-1}\right]\)
• C. \(y\left[\ln(x+1)-\ln(x-1)\right]\)
• D. \(\dfrac{2y}{x^2-1}\)

Hint:

Whenever both the base and exponent contain variables, logarithmic differentiation converts complicated exponential expressions into manageable logarithmic forms.

Answer 1

The Correct Option is A

Concept: Whenever a variable appears simultaneously in both the base and exponent, ordinary differentiation becomes complicated. In such situations, logarithmic differentiation is the most efficient and systematic method. It converts exponential expressions into logarithmic forms where product rule and implicit differentiation can be applied easily.

Step 1: Identifying the structure of the function.
Given: \[ y = \left(\dfrac{x+1}{x-1}\right)^x \] Observe carefully that:
• The base contains the variable \(x\)
• The exponent also contains the variable \(x\) Thus the function is of the type: \[ [f(x)]^{g(x)} \] Hence logarithmic differentiation must be used.

Step 2: Taking natural logarithm on both sides.
Applying logarithm: \[ \ln y = \ln\left[ \left(\dfrac{x+1}{x-1}\right)^x \right] \] Using logarithmic identity: \[ \ln(a^b)=b\ln a \] we get: \[ \ln y = x\ln\left(\dfrac{x+1}{x-1}\right) \] Now split the logarithm: \[ \ln\left(\dfrac{x+1}{x-1}\right) = \ln(x+1)-\ln(x-1) \] Thus, \[ \ln y = x[\ln(x+1)-\ln(x-1)] \]

Step 3: Differentiating both sides implicitly.
Differentiate with respect to \(x\). Left-hand side: \[ \dfrac{d}{dx}(\ln y) = \dfrac{1}{y}\dfrac{dy}{dx} \] Now differentiate the right-hand side: \[ \dfrac{d}{dx} \left[ x\ln\left(\dfrac{x+1}{x-1}\right) \right] \] Using product rule: \[ \dfrac{d}{dx}(uv) = u\dfrac{dv}{dx} + v\dfrac{du}{dx} \] Take: \[ u=x \] and \[ v=\ln\left(\dfrac{x+1}{x-1}\right) \] Therefore, \[ \dfrac{1}{y}\dfrac{dy}{dx} = x\cdot \dfrac{d}{dx} \left[ \ln\left(\dfrac{x+1}{x-1}\right) \right] + \ln\left(\dfrac{x+1}{x-1}\right)\cdot1 \] Rearranging: \[ \dfrac{1}{y}\dfrac{dy}{dx} = \ln\left(\dfrac{x+1}{x-1}\right) + x \dfrac{d}{dx} \left[ \ln\left(\dfrac{x+1}{x-1}\right) \right] \]

Step 4: Differentiating the logarithmic term.
Using: \[ \ln\left(\dfrac{x+1}{x-1}\right) = \ln(x+1)-\ln(x-1) \] Differentiate term by term: \[ \dfrac{d}{dx} \left[ \ln(x+1)-\ln(x-1) \right] = \dfrac{1}{x+1} - \dfrac{1}{x-1} \] Now simplify: \[ \dfrac{1}{x+1} - \dfrac{1}{x-1} = \dfrac{x-1-(x+1)}{(x+1)(x-1)} \] \[ = \dfrac{x-1-x-1}{x^2-1} \] \[ = \dfrac{-2}{x^2-1} \] Thus, \[ \dfrac{d}{dx} \left[ \ln\left(\dfrac{x+1}{x-1}\right) \right] = \dfrac{-2}{x^2-1} \]

Step 5: Substituting back into derivative expression.
Substitute into the previous equation: \[ \dfrac{1}{y}\dfrac{dy}{dx} = \ln\left(\dfrac{x+1}{x-1}\right) + x\left( \dfrac{-2}{x^2-1} \right) \] \[ \dfrac{1}{y}\dfrac{dy}{dx} = \ln\left(\dfrac{x+1}{x-1}\right) - \dfrac{2x}{x^2-1} \]

Step 6: Obtaining the final answer.
Multiply both sides by \(y\): \[ \dfrac{dy}{dx} = y \left[ \ln\left(\dfrac{x+1}{x-1}\right) - \dfrac{2x}{x^2-1} \right] \] Therefore, \[ \boxed{(A)\ \dfrac{dy}{dx} = y \left[ \ln\left(\dfrac{x+1}{x-1}\right) - \dfrac{2x}{x^2-1} \right] } \]