Question

if x^y +y^x = (x+y)^x+y then find dy/dx​

Answer 1

Answer:    $\frac{dy}{dx} = \frac{(x+y)^{x+y}[1+log(x+y)]-y[x^{y-1}+y^{x-1}logy]}{x[x^{y-1}logx+y^{x-1}]-(x+y)^{x+y}[1+log(x+y)]}$

Step-by-step explanation:

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

assume $u=x^{y}$ , $v=y^{x}$ , $t=(x+y)^{x+y}$

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

$\frac{du}{dx} +\frac{dv}{dx} = \frac{dt}{dx}$

$u=x^{y}\\logu=log(x^{y})\\\frac{d(logu)}{dx} = \frac{d(log(x^{y}))}{dx} \\\frac{d(logu)}{dx} = \frac{d(ylog(x))}{dx} \\\frac{1}{u} \frac{du}{dx} = \frac{y}{x} + log(x )\frac{dy}{dx} \\\frac{du}{dx} = u [\frac{y}{x} + log(x )\frac{dy}{dx} ]$

$v=y^{x} \\logv=log(y^{x})\\logv=xlog(y)\\\frac{d(logv)}{dx}=\frac{d(xlog(y))}{dx} \\\frac{1}{v} \frac{dv}{dx} = logy+\frac{x}{y} \frac{dy}{dx} \\\frac{dv}{dx} =v[ logy+\frac{x}{y} \frac{dy}{dx} ]$

$t=(x+y)^{x+y}\\logt=(x+y)log(x+y)\\\frac{1}{t} \frac{dt}{dx} = \frac{x+y}{x+y}(1+\frac{dy}{dx} )+(1+\frac{dy}{dx} )log(x+y)\\\frac{1}{t} \frac{dt}{dx}=(1+\frac{dy}{dx} )+(1+\frac{dy}{dx} )log(x+y)\\\frac{dt}{dx}=t[1+log(x+y)][1+\frac{dy}{dx}]\\\frac{dt}{dx}=(x+y)^{x+y}[1+log(x+y)][1+\frac{dy}{dx}]$

$u[\frac{y}{x}+log(x)\frac{dy}{dx}]+v[logy+\frac{x}{y} \frac{dy}{dx}]=(x+y)^{x+y} [1+log(x+y)][1+\frac{dy}{dx}]$

$x^{y} [\frac{y}{x}+log(x)\frac{dy}{dx}]+y^{x} [logy+\frac{x}{y} \frac{dy}{dx}]=(x+y)^{x+y} [1+log(x+y)][1+\frac{dy}{dx}]$

$\frac{dy}{dx} = \frac{(x+y)^{x+y}[1+log(x+y)]-y[x^{y-1}+y^{x-1}logy]}{x[x^{y-1}logx+y^{x-1}]-(x+y)^{x+y}[1+log(x+y)]}$