Cho hai hàm số f(x)=tanx  và g(x)=$\frac{1}{1-x}. Tính \frac{f\text{'}\left(0\right)}{g\text{'}\left(0\right)}$

Giải

$$\begin{gathered} f\text{'}\left(x\right)=\frac{1}{{\cos }^{2}x}\Rightarrow f\text{'}\left(0\right)=1; g\text{'}\left(x\right)=\frac{1}{{\left(1-x\right)}^2}\Rightarrow g\text{'}\left(0\right)=1 \\ Vậy \frac{f\text{'}\left(0\right)}{g\text{'}\left(0\right)}=1 \end{gathered}$$