A calculus problem by Rajdeep Dhingra

$let\quad f\left( x \right) =y\\ so\quad f\left( x \right) =\quad { x }^{ y }\\ \quad \quad \quad y={ x }^{ y }\\ take\quad log\quad both\quad side\\ log\quad y\quad =\quad y*log\quad x\\ now\quad differciating\quad both\quad sides\quad we\quad get\\ \frac { 1 }{ y } \frac { dy }{ dx } \quad =\quad \frac { y }{ x } +\quad \frac { dy }{ dx } (log\quad x)\\ on\quad solving\quad \\ \frac { dy }{ dx } =\quad \frac { y }{ x[\frac { 1 }{ y } -\quad log\quad x] } \\ putting\quad x=1\quad (\quad y\quad will\quad also\quad be\quad 1\quad )\quad we\quad get\\ \frac { d }{ dx } [f\left( x \right)]=\quad 1$

f'(x)=x(x^x^x^x^(x-1)=x[f(x)] f'(1)=(1)[f(1)] = 1

Let f(x) = y =x^y Taking log

ln(y) = y ln(x)

Y'=y^2/[x{1-ln(y)]

x=1,y=1 Putting we get

Y'=1