The problem asks to find $y'$ (the derivative of $y$ with respect to $x$) for two equations: a) $x^y = y^x$ b) $(\cosh(x))^y = y^{\cosh(x)}$

1. Problem Description

The problem asks to find $y'$ (the derivative of $y$ with respect to $x$) for two equations:

a) $x^y = y^x$

b) $(\cosh(x))^y = y^{\cosh(x)}$

2. Solution Steps

a) $x^y = y^x$

Take the natural logarithm of both sides:

$y \ln(x) = x \ln(y)$

Differentiate both sides with respect to $x$ using the product rule:

$\frac{d}{dx} (y \ln(x)) = \frac{d}{dx} (x \ln(y))$

$y' \ln(x) + y \cdot \frac{1}{x} = 1 \cdot \ln(y) + x \cdot \frac{1}{y} \cdot y'$

$y' \ln(x) + \frac{y}{x} = \ln(y) + \frac{x}{y} y'$

$y' \ln(x) - \frac{x}{y} y' = \ln(y) - \frac{y}{x}$

$y' (\ln(x) - \frac{x}{y}) = \ln(y) - \frac{y}{x}$

$y' = \frac{\ln(y) - \frac{y}{x}}{\ln(x) - \frac{x}{y}}$

$y' = \frac{\frac{x \ln(y) - y}{x}}{\frac{y \ln(x) - x}{y}}$

$y' = \frac{y(x \ln(y) - y)}{x(y \ln(x) - x)}$

Since $x^y = y^x$, then $y \ln(x) = x \ln(y)$.

$y' = \frac{y(x \ln(y) - y)}{x(x \ln(y) - x)}$

$y' = \frac{y(y \ln(x) - y)}{x(y \ln(x) - x)}$

$y' = \frac{y}{x} \frac{y \ln(x) - y}{y \ln(x) - x}$

$y' = \frac{y}{x} \frac{y(\ln(x) - 1)}{y \ln(x) - x}$

$y' = \frac{y}{x} \frac{y (\ln x - 1)}{x \ln y - x}$

$y' = \frac{y}{x} \frac{y (\ln x - 1)}{x (\ln y - 1)}$

$y' = \frac{y}{x} \frac{y(\ln(x) - 1)}{x(\ln(y) - 1)}$

Let's simplify

$y' = \frac{\ln(y) - \frac{y}{x}}{\ln(x) - \frac{x}{y}} = \frac{x\ln(y) - y}{x} \cdot \frac{y}{y\ln(x) - x} = \frac{y(x\ln(y) - y)}{x(y\ln(x) - x)}$

Since $x^y = y^x$, $x\ln(y) = y\ln(x)$. Let $A = x\ln(y) = y\ln(x)$.

$y' = \frac{y(A - y)}{x(A - x)}$

b) $(\cosh(x))^y = y^{\cosh(x)}$

Take the natural logarithm of both sides:

$y \ln(\cosh(x)) = \cosh(x) \ln(y)$

Differentiate both sides with respect to $x$:

$y' \ln(\cosh(x)) + y \frac{\sinh(x)}{\cosh(x)} = \sinh(x) \ln(y) + \cosh(x) \frac{1}{y} y'$

$y' \ln(\cosh(x)) + y \tanh(x) = \sinh(x) \ln(y) + \frac{\cosh(x)}{y} y'$

$y' \ln(\cosh(x)) - \frac{\cosh(x)}{y} y' = \sinh(x) \ln(y) - y \tanh(x)$

$y' (\ln(\cosh(x)) - \frac{\cosh(x)}{y}) = \sinh(x) \ln(y) - y \tanh(x)$

$y' = \frac{\sinh(x) \ln(y) - y \tanh(x)}{\ln(\cosh(x)) - \frac{\cosh(x)}{y}}$

$y' = \frac{y(\sinh(x) \ln(y) - y \tanh(x))}{y \ln(\cosh(x)) - \cosh(x)}$

3. Final Answer

a)

$y' = \frac{y(x \ln(y) - y)}{x(y \ln(x) - x)}$

b)

$y' = \frac{y(\sinh(x) \ln(y) - y \tanh(x))}{y \ln(\cosh(x)) - \cosh(x)}$