Let y be an implicit function of x defined by x2x−2xxcot⁡y−1=0  Then y1(1) equals

put x=1⇒y=π2 and then differentiate u=x2xdudx=x2x(2log⁡x+2)v=xxdvdx=xx(1+log⁡x)x2x−2xxcot⁡y−1=0 Differentiation w.r.to ‘x’x2x(2log⁡x+2)−2xx(1+log⁡x)cot⁡y+2xxcosec2⁡yy1=0⇒1(2(0)+2)−2(1+0)(0)+2y1=0⇒2+2y1=0⇒y1=−22⇒y1=−1