Question

Pls solve question 32

Answer 1

Solution

let ...

$x {}^{y} = u \: \: and \: \: y {}^{x} = v$

now...

$x {}^{y} + y {}^{x} = a {}^{b} \\ = > u + v = a {}^{b} \\ = > \frac{du}{dx} + \frac{dv}{dx} = 0$

now...

$u = x {}^{y} \\ = > log(u) = y log(x) \\ = > \frac{1}{u} \frac{du}{dx} = \frac{y }{x} + log(x) \frac{dy}{dx} \\ = > \frac{du}{dx} = yx {}^{y - 1} + x {}^{y} log(x) \frac{dy}{dx}$

and...

$v = y {}^{x} \\ = > log(v) = x log(y) \\ = > \frac{1}{v} \frac{dv}{dx} = \frac{x}{y} \frac{dy}{dx} + log(y) \\ = > \frac{dv}{dx} = xy {}^{x - 1} \frac{dy}{dx} + y {}^{x} log(y)$

therefore....

$= > \frac{du}{dx} + \frac{dv}{dx} = 0 \\ = > yx {}^{y - 1} +x {}^{y} log(x) \frac{dy}{dx} + xy {}^{x - 1} \frac{dy}{dx} + y {}^{x} log(y) = 0 \\ = > \frac{dy}{dx} ( {x}^{y} log(x) + xy {}^{x - 1} ) = - (yx {}^{y - 1} + y {}^{x} log(y) ) \\ = > \frac{dy}{dx} = \frac{ - (yx {}^{y - 1} + y {}^{x} log(y) )}{( \: x {}^{y} log(x) + xy {}^{x - 1}) }$

Hope this helps you.....✌️