If `y=[2x^3+3][2x^-3+1]`, then find `(dy)/(dx)`.

Options

A. `-18/x^3+6x^3`
B. `-9/x^3+6x^2`
C. `-18/x^4+6x^2`
D. `-1/x^4+3x^2`

Answer

C

Explanation

Here, `u=2x^3+3`, `v=2x^-3+1`.
Differentiating both sides w.r.t. x, we get
`(dy)/(dx)=(d)/(dx)[(2x^3+3)(2x^-3+1)]`
Using product rule, `(d(uv))/(dx)=u(dv)/(dx)+v(du)/(dx)`
`(dy)/(dx)=[2x^3+3](d)/(dx)[2x^-3+1]+[2x^-3+1](d)/(dx)[2x^3+3]`
`=[2x^3+3][-6x^-4]+[2x^-3+1][6x^2]`
`=[2x^3+3][(-6)/(x^4)]+[2/x^3+1][6x^2]=-18/x^4+6x^2`