Question

54. Find $y^{\prime}$ if $x^y=y^x\text{.}$

Answer 1

Step-by-step explanation:

Take the natural log of both sides:

$ln ({x}^{y} ) = ln ({y}^{x} )$

Logarithm rules allow you to bring down the exponents:

$yln(x) = xln(y)$

Now differentiate. We will have to implicitly differentiate 'y' since it is a function of 'x'. Both sides require the product rule:

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

Isolate the terms that have y' since that is what we want:

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

Factor out y' to get:

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

Therefore:

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