Question

How do you differentiate `y = x^cos(3x)`?

Answer 1

`dy/dx= (cos(3x)/x -3sin(3x)lnx)x^cos(3x)`

Explanation:

Take logs of both sides:

`y = x^cos(3x)`
`:. lny = ln(x^cos(3x))`
`:. lny = cos(3x)lnx`

The LHS can be differentiated simplicity, and the RHS can be differentiated using product rule; `d/dx(uv)=u(dv)/dx+v(du)/dx`

So, we have:
`1/ydy/dx= cos(3x)d/dxlnx + lnxd/dxcos(3x)`
`:. 1/ydy/dx= cos(3x)1/x + (lnx)(-sin(3x))(3)`
`:. 1/ydy/dx= cos(3x)/x -3sin(3x)lnx`
`:. dy/dx= (cos(3x)/x -3sin(3x)lnx)y`
`:. dy/dx= (cos(3x)/x -3sin(3x)lnx)x^cos(3x)`