Question

x^y + y^x+x^x = 2 Find dy/dx​

Answer 1

Question:

${x}^{y} + {y}^{x} + {x}^{x} = 2$

Answer:

$putting \: {x}^{y} = u \: \: {y}^{x} = v \: \: {x}^{x} = w \\ we \: get \\ u + v + w = 2$

differentiate w.r.t.x

$\frac{du}{dx} + \frac{dv}{dx} + \frac{dw}{dx} = 0 \\ \\ now \: u = {x}^{y} \\ \\ taking \: log \: on \: both \: side \\ \\ log \: u = \: log {x}^{y} \implies \: logu = ylogx \\ \\ diff \: w.r.t.x \\ \\ \implies \frac{1}{u} . \frac{du}{dx} = y. \frac{1}{x} + logx \: \frac{dy}{dx} \\ \\ \implies \frac{du}{dx} = u( \frac{y}{x} + logx \frac{dy}{dx} ) \\ \\ \implies \frac{du}{dx} = {x}^{y} ( \frac{y}{x} + logx \frac{dy}{dx} )..............(2)$

Taking logs

$logv = log {y}^{x} \\ \\ \implies \: logv = xlogy \\ \\ diff \: w.r.t.x \\ \\ \implies \frac{1}{v} \frac{dv}{dx} = x \frac{ 1}{y} \frac{dy}{dx} + logy.1 \\ \\ \implies \: \frac{dv}{dx} = v( \frac{x}{y} \frac{dy}{dx} + logy) \\ \\ \implies \: \frac{dv}{dx} = {y}^{x} ( \frac{x}{y} \frac{dy}{dx} + logy)........(3)$

Lastly

$w = {x}^{x} \\ \\ \therefore \: \frac{dw}{dx} = {x}^{x} (1 + logx).........(4)$

from 1. using (1),(2),(3) we get......

${x}^{y} (\frac{y}{x} + logx \frac{dy}{dx} ) + {y}^{x} ( \frac{x}{y} \frac{dy}{dx} + logy) + {x}^{x} 1 + logx) = 0 \\ \\ \implies \:( xy ^{x - y} + {x}^{y} logx) \frac{dy}{dx} = - {x}^{x} (1 + logx) - y {x}^{y - 1} - {y}^{x} logy \\ \\ hence \\ \\ \frac{dy}{dx} = - \frac{ {y}^{x}logy + y. {x}^{y - 1} + {x}^{x} (1 + logx) }{x( {y}^{x - 1} + {x}^{y - 1}logx }$