${x^{2x}} - 2{x^x}\,\cot \,y - 1 = 0$ $\Rightarrow 2\,\cot \,y = {x^x} - {x^{ - x}}$ $\Rightarrow 2\,\cot \,y\, = u - {1 \over u}$ where $u = {x^x}$ Differentiating both sides with respect to $x,$ we get $\Rightarrow - 2\cos e{c^2}y{{dy} \over {dx}}$ $= \left( {1 + {1 \over {{u^2}}}} \right){{du} \over {dx}}$ where $u = {x^x} \Rightarrow \log \,u = x\,\log \,x$ $\Rightarrow {1 \over u}{{du} \over {dx}} = 1 + \log \,x$ $\Rightarrow {{du} \over {dx}} = {x^x}\left( {1 + \log \,x} \right)$ $\therefore$ We get $- 2\cos e{c^2}y{{dy} \over {dx}}$ $= \left( {1 + {x^{ - 2x}}} \right){x^x}\left( {1 + \log \,x} \right)$ $\Rightarrow {{dy} \over {dx}} = {{\left( {{x^x} + {x^{ - x}}} \right)\left( {1 + \log x} \right)} \over { - 2\left( {1 + {{\cot }^2}y} \right)}}\,\,\,\,\,\,...\left( i \right)$ Now when $x=1,$ ${x^{2x}} - 2{x^x}\,\cot \,y - 1 = 0,$ gives $1 - 2\,\cot y - 1 = 0$ $\Rightarrow \,\,\cot y\, = 0$ $\therefore$ From equation $(i),$ at $x=1$ and $\cot \,y = 0,$ we get $y'\left( 1 \right) = {{\left( {1 + 1} \right)\left( {1 + 0} \right)} \over { - 2\left( {1 + 0} \right)}} = - 1$