#### Answer

$ f(x)=g(x)=x $
$ f'(x)=g'(x)=\frac{d}{dx}(x)=1$
$(\frac{f}{g})'=\frac{f'g-fg'}{g^{2}}=\frac{1\times x-x\times1}{x^{2}}=0$
$\frac{f'}{g'}=\frac{1}{1}=1$
We see that $(\frac{f}{g})'\ne \frac{f'}{g'}$

#### Work Step by Step

See the answer above.