Question

`(sinx)^(logx)`

Answer 1

Let `y = (sinx)^logx`
Taking logs both sides,
`logy = logx logsin x `
Differentiating both sides w.r.t. x
`1/y dy/dx = logx(1/sinx)(cosx)+ logsinx (1/x)`
`=>1/ydy/dx = logxcotx+logsinx/x`
`=> dy/dx = y( logxcotx+logsinx/x)`
`=> dy/dx = (sinx)^logx( logxcotx+logsinx/x)`, which is the required solution.