Question

dy/dx of (logx) to the power cosx

Answer 1

let y=logx^cosx
applying log to both sides-
Log y = log (logx)^cosx
logy = cosx. log (logx)
Differentiating+
1/y dy/dx = d/dx {cosx. (1/logx)}
dy/dx = y. cosx.x +sinx. 1/logx
= logx^cosx {cosx.x+sinx (1/logx)}