Question

Find `dy/dx` of `y=sin(cx)sin^c(x)`?

Answer 1

`dy/dx=csin(cx)cos(x)sin^(c-1)(x)+csin^c(x)cos(cx)=csin(x)^(c-1)sin(cx+x)`

Explanation:

For a given function `y=f(x)=uv` where `u` and `v` are both functions of `x` we get:
`dy/dx=u'v+v'u`

`u=sin(cx)`
`u'=c cos(cx)`

`v=sin^c(x)`
`v'=c cos(x)sin^(c-1)(x)`

`dy/dx=csin(cx)cos(x)sin^(c-1)(x)+csin^c(x)cos(cx)=csin(x)^(c-1)sin(cx+x)`