与 [CJC/2013/Promo/4]

1. Simplify the argument of the logarithm and apply the laws of logarithms to expand the expression.

$\begin{aligned} \ln \sqrt{\frac{(x+1)^3}{x^2 - 1}} &= \frac{1}{2} \ln \left( \frac{(x+1)^3}{(x+1)(x-1)} \right) \\[1ex] &= \frac{1}{2} \ln \left( \frac{(x+1)^2}{x-1} \right) \\[1ex] &= \frac{1}{2} [ 2\ln(x+1) - \ln(x-1) ] \\[1ex] &= \ln(x+1) - \frac{1}{2}\ln(x-1) \end{aligned}$

2. Differentiate the expanded expression with respect to $x$.

$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x} \left[ \ln(x+1) - \frac{1}{2}\ln(x-1) \right] &= \frac{1}{x+1} - \frac{1}{2(x-1)} \end{aligned}$

3. Combine the resulting terms into a single algebraic fraction.

$\begin{aligned} \frac{1}{x+1} - \frac{1}{2(x-1)} &= \frac{2(x-1) - (x+1)}{2(x+1)(x-1)} \\[1ex] &= \frac{2x - 2 - x - 1}{2(x^2 - 1)} \\[1ex] &= \frac{x-3}{2(x^2-1)} \end{aligned}$

1. Apply the chain rule and the standard derivative of the inverse tangent function.

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{1 + (\sqrt{x})^2} \cdot \frac{\mathrm{d}}{\mathrm{d}x}(\sqrt{x})$

2. Evaluate the derivative of $\sqrt{x}$ and simplify the final expression.

$\frac{\mathrm{d}}{\mathrm{d}x}(\sqrt{x}) = \frac{1}{2}x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}}$

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{1 + x} \cdot \frac{1}{2\sqrt{x}}$

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2\sqrt{x}(1 + x)}$

1. Express the relation using fractional indices, take the natural logarithm on both sides, and simplify.

$y^{\frac{1}{x}} = x^{\frac{1}{y}}$
$\ln\left(y^{\frac{1}{x}}\right) = \ln\left(x^{\frac{1}{y}}\right)$
$\frac{1}{x} \ln y = \frac{1}{y} \ln x$
$y \ln y = x \ln x$

2. Differentiate implicitly with respect to $x$ using the product rule.

$\frac{\mathrm{d}}{\mathrm{d}x}(y \ln y) = \frac{\mathrm{d}}{\mathrm{d}x}(x \ln x)$
$\frac{\mathrm{d}y}{\mathrm{d}x} \ln y + y \left( \frac{1}{y} \frac{\mathrm{d}y}{\mathrm{d}x} \right) = 1 \cdot \ln x + x \left( \frac{1}{x} \right)$
$\frac{\mathrm{d}y}{\mathrm{d}x} \ln y + \frac{\mathrm{d}y}{\mathrm{d}x} = \ln x + 1$

3. Factorise and isolate $\frac{\mathrm{d}y}{\mathrm{d}x}$ to find the final derivative.

$\frac{\mathrm{d}y}{\mathrm{d}x}(\ln y + 1) = \ln x + 1$
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\ln x + 1}{\ln y + 1}$