Question

y=x^y + y^x +x^x= a^b

dy/dx =?​

Answer 1

Answer:

Step-by-step explanation:

$a^b=x^y+y^x+x^x\\\\let \\\\a=x^y\\\\b=y^x\\\\c=x^x$

$\frac{d(a^b)}{dx}=\frac{da}{dx}+\frac{db}{dx}+\frac{dc}{dx}$

$a=x^y\\apply\;\;log_e\\\\log_ea=log_ex^y\\\\loga=ylogx\\\\diff\;\;w.r.t.\;\;x\\\\\frac{1}{a}\frac{da}{dx}=logx\frac{dy}{dx}+\frac{y}{x}\\\\\frac{da}{dx}=a(logx\frac{dy}{dx}+\frac{y}{x})\\\\\frac{da}{dx}=x^y(logx\frac{dy}{dx}+\frac{y}{x})$

$b=y^x\\apply\;log_e\\log_eb=log_ey^x\\\\logb=xlogy\\\\diff\;\;w.r.t.\;\;x\\\\\frac{1}{b}\frac{db}{dx}=logy+\frac{x}{y}\frac{dy}{dx}\\\\\frac{db}{dx}=b(logy+\frac{x}{y}\frac{dy}{dx})\\\\\frac{db}{dx}=y^x(logy+\frac{x}{y}\frac{dy}{dx})$

$c=x^x\\apply\;\;log_e\\log_ec=log_ex^x\\\\logc=xlogx\\\\diff\;\;w.r.t.\;\;x\\\\\frac{1}{c}\frac{dc}{dx}=logx+1\\\\\frac{dc}{dx}=c(logx+1)\\\\\frac{dc}{dx}=x^x(1+logx)$

$\large{\bold{BUT\;\;\frac{d(a^b)}{dx}=0}}\\\\\large{\bold{=>\;\;\frac{da}{dx}+\frac{db}{dx}+\frac{dc}{dx}=0}}$

$x^y(logx\frac{dy}{dx}+\frac{y}{x})+y^x(logy+\frac{x}{y}\frac{dy}{dx})+x^x(1+logx)=0\\\\\frac{dy}{dx}(x^ylogx+xy^{x-1})+x^{y-1}y+y^xlogy+x^x+x^xlogx=0\\\\\frac{dy}{dx}(x^ylogx+xy^{x-1})=-(x^{y-1}y+y^xlogy+x^x+x^xlogx)\\\\\frac{dy}{dx}=-\frac{(x^{y-1}y+y^xlogy+x^x+x^xlogx)}{(x^ylogx+xy^{x-1})}\\\\\large{\bold{\frac{dy}{dx}=-\frac{(x^{y-1}y+y^xlogy+x^x+x^xlogx)}{(x^ylogx+xy^{x-1})}}}$

Hope it Helps