`y=(x^2+1)sin^-1 4x`

Let `u=x^2+1` and `v=sin^(-1) 4x`.

Then

`(dy)/(dx)=u (dv)/(dx)+v (du)/(dx)`

`=(x^2+1)[1/(sqrt(1-(4x)^2)) (4)]+` `(sin^(-1) 4x)(2x)`

`=(4(x^2+1))/(sqrt(1-16x^2))+2x\ sin^(-1)4x`

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