Calculus (3rd Edition)

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