Question

y= cos (logx)/x then dy/dx ??​

Answer 1

Answer:

y = ( cos x)log x + (log x )x

tahing log on both sides

log y = log x .log( cos x) +x . log ( log x)

differentiating ;

1/y dy/dx = log (cos x)/x - tan x log x +log ( log x ) +1/log x

dy/dx = y[log ( cos x )/ x - tan x log x +log( log x ) +1/log x ]

therefore , dy/dx = ( cos x )log x + ( log x )x [ log (cos x )/ x -tan x log x +log ( log x ) +1/log x