use logarithmic differentiation to find $\frac{dy}{dx}$ for $y=(1 + x)^{\frac{2}{x}}$.

Explanation:

Step1: Take natural - log of both sides

$\ln y=\ln(1 + x)^{\frac{2}{x}}=\frac{2}{x}\ln(1 + x)$

Step2: Differentiate both sides with respect to (x)

Using the product rule ((uv)^\prime = u^\prime v+uv^\prime) where (u=\frac{2}{x}) and (v = \ln(1 + x)). The derivative of (u=\frac{2}{x}=2x^{-1}), so (u^\prime=- 2x^{-2}=-\frac{2}{x^{2}}). The derivative of (v=\ln(1 + x)) is (v^\prime=\frac{1}{1 + x}). (\frac{1}{y}\frac{dy}{dx}=-\frac{2}{x^{2}}\ln(1 + x)+\frac{2}{x}\cdot\frac{1}{1 + x})

Step3: Solve for (\frac{dy}{dx})

Multiply both sides by (y=(1 + x)^{\frac{2}{x}}) (\frac{dy}{dx}=(1 + x)^{\frac{2}{x}}\left(\frac{2}{x(1 + x)}-\frac{2\ln(1 + x)}{x^{2}}\right))

Answer:

((1 + x)^{\frac{2}{x}}\left(\frac{2}{x(1 + x)}-\frac{2\ln(1 + x)}{x^{2}}\right))