Let f(x) and g(x) be two functions having finite non-zero 3^rd order derivatives f'''(x) and g'''(x) for all x ∈ R. If f(x) g(x) = 1 for all x ∈ R, then $\frac{f'''}{f'}-\frac{g'''}{g'}$ is equal to

$3\left(\frac{f''}{f}-\frac{g''}{g}\right)$

We have,

$f(x) g(x)=1$

Differentiating with respect to x, we get

$f' g+f g'=0$            ...........(i)

Differentiating (i) w.r.t. x, we get

$f'' g+2 f' g'+f g''=0$            ...........(ii)

Differentiating (ii) w.r.t. x, we get

$f''' g+g''' f+3 f'' g'+3 g'' f'=0$

$\Rightarrow \frac{f'''}{f'}\left(f' g\right)+\frac{g'''}{g'}\left(f g'\right)+\frac{3 f''}{f}\left(f g'\right)+\frac{3 g''}{g}\left(g f'\right)=0$

$\Rightarrow\left(\frac{f'''}{f'}+\frac{3 g''}{g}\right)\left(f' g\right)=-\left(\frac{g'''}{g'}+\frac{3 f''}{f}\right)\left(f g'\right)$

$\Rightarrow-\left(\frac{f'''}{f'}+\frac{3 g''}{g}\right)\left(f g'\right)=-\left(\frac{g'''}{g'}+\frac{3 f''}{g}\right) f g'$           [Using (1)]

$\Rightarrow \frac{f'''}{f'}+\frac{3 g''}{g}=\frac{g'''}{g'}+\frac{3 f''}{f}$

$\Rightarrow \frac{f'''}{f'}-\frac{g'''}{g'}=3\left(\frac{f''}{f}-\frac{g''}{g}\right)$