Suppose `fa n dg` are functions having second derivative `f"` and `g' '` everywhere. If `f(x)dotg(x)=1` for all `xa n df^(prime)a n dg'` are never zero, then `(f^(x))/(f^(prime)(x))-(g^(x))/(g^(prime)(x))e q u a l` `(-2f^(prime)(x))/f` (b) `(2g^(prime)(x))/(g(x))` `(-f^(prime)(x))/(f(x))` (d) `(2f^(prime)(x))/(f(x))`

Options

A. `(-2f'(x))/(f(x))`
B. `(-2g'(x))/(g(x))`
C. `(-f'(x))/(f(x))`
D. `(2f'(x))/f(x)`

Explanation

`"We have "=(1)/(f)`
`therefore" "g'=(-1)/(f^(2))f'`
`"or "g''=-[-(2)/(f^(3))f'^(2)+(1)/(f^(2))f'']`
`=(2)/(f^(3))f'^(2)-(f'')/(f^(2))`
`"or "(f'')/(f')-(g'')/(g')=(f'')/(f')-((2)/(f^(3))f'^(2)-(f'')/(f^(2)))/(-(1)/(f^(2))f')`
`=(f'')/(f')-((-2f')/(f)+(f'')/(f'))=(2f')/(f)`
`"Alos, "g cdotf=1`
`"or "g'f+gf'=0`
`therefore" "(f'')/(f')-(g'')/(g')=-(2g')/(g)`