Question

differentiate y w.r.t. x where y=y^(cot^-1x^4)^7

Answer 1

y=y^(cot-1x^4)^7
taking log on both side
ln y = ln y^cot^-1x^4)^7
lny=(cot^-1x^4)lny
1/y dy/dx =  (cot^-1x^4 )^7. d/dx  lny + y .d/dx (cot^-1x^4)^7
1/y dy/dx = (cot^-1x^4)^7 . 1/y dy/dx  + y ( 7 cot-x^4)^6 . d/dx (cot ^-x^4)
1/y dy/dx=(cot^-1x^4)^7.1/y dy/dx + y(7 cot^-x^4)^6. (-1/1+(x^4)^2 . d/dx x^4
1/y dy/dx =(cot ^-1 x^4)^7 . 1/y dy/dx + y(7 cot^-x^4)^6(-1/1+(x^4)^2).4x^3
1/y . dy/dx - 1/y.dy/dx (cot^-1x^4)^7= y(7 cot^-x^4)^6(-1/1+(x^4)^2).4x^3
dy/dx{1/y - 1/y.dy/dx (cot^-1x^4)^7} = y(7 cot^-x^4)^6(-1/1+(x^4)^2).4x^3
dy/dx =[y(7 cot^-x^4)^6(-1/1+(x^4)^2).4x^3]÷{1/y - 1/y.dy/dx (cot^-1x^4)^7}
hence find this differentation toooooooooo  hard  finally i solve thi :)