## Calculus 10th Edition

Counter-example: $f(x)=(x-1)(x+1)\rightarrow g(x)=(x-1)$; $h(x)=(x+1)$ $g'(x)=1$; $h'(x)=1\rightarrow$ If it was true $f'(x)=1$ But by expanding we see that $f(x)=x^2-1\rightarrow f'(x)=2x.$ The rule is therefore false. Also, it contradicts the product rule.