how do you derive x^sinx?

Answer 1

let x^sinx be y
taking log
ln y=sinx ln x
1/y *dy/dx=sinx/x+cosxlnx
dy/dx=y[(sinx/x+lnxcosx]
=x^sinx[(sinx/x+lnxcosx]

Answer 2

Solution:-
Let y=x^sinx--------------------- (1)
Taking log on both the sides,
log y=log x^sinx
Therefore, log y=sinx.logx
Differentiating w.r.t.x,
1/y. dy/dx=sinx.1/x+logx.cosx (Using product rule)
dy/dx=y(sinx/x+logx.cosx)
dy/dx=x^sinx(sinx/x+logx.cosx)----- {from statement no1}

Answer 3

Let y=x^sinx
=>log y = xlog(sin(x))
=>(1/y)dy/dx=x(1/sin(x))*cosx + log(sin(x))...by chain rule!
=>dy/dx=[xcot(x) + log(sin(x))]x^sin(x)