If `u=f(x^3),v=g(x^2),f^(prime)(x)=cosx ,a n dg^(prime)(x)=sinx ,t h e n(d u)/(d v)` is `3/2xcosx^3cos e cx^2` `2/3sinx^3secx^2` `tanx` (d) none of these

Options

A. `(3)/(2)x " cos "x^(2) " cosec "x^(2)`
B. `(3)/(2)" sin "x^(3) " sec "x^(2)`
C. `tan x`
D. none of these

Explanation

`(du)/(dv)=((du)/(dx))/((dv)/(dx))=(f'(x^(3))3x^(2))/(g'(x^(2))2x)=(cos x^(3)(3x^(2)))/(sin x^(2)(2x))=(3)/(2)x cos x^(3) cosec x^(2)`