Cho hàm số $f\left(x\right)=\frac{x}{\sqrt{4-x^2}}$  và $g\left(x\right)=\frac{1}{x}+\frac{1}{\sqrt{x}}+x^2$ . Tính f'(0) – g'(1).

Có $f^{\text{'}}\left(x\right)={\left(\frac{x}{\sqrt{4-x^2}}\right)}^{\text{'}}$ $=\frac{x^{\text{'}}\cdot \sqrt{4-x^2}-x\cdot {\left(\sqrt{4-x^2}\right)}^{\text{'}}}{4-x^2}$

$=\frac{\sqrt{4-x^2}-x\cdot \frac{-2x}{2\sqrt{4-x^2}}}{4-x^2}$$=\frac{\sqrt{4-x^2}+\frac{x^2}{\sqrt{4-x^2}}}{4-x^2}$

$=\frac{4-x^2+x^2}{\left(4-x^2\right)\sqrt{4-x^2}}$$=\frac{4}{\left(4-x^2\right)\sqrt{4-x^2}}$

Khi đó $f^{\text{'}}\left(0\right)=\frac{4}{\left(4-0\right)\sqrt{4-0}}=\frac{1}{2}$

Có $g^{\text{'}}\left(x\right)={\left(\frac{1}{x}+\frac{1}{\sqrt{x}}+x^2\right)}^{\text{'}}$$={\left(\frac{1}{x}\right)}^{\text{'}}+{\left(\frac{1}{\sqrt{x}}\right)}^{\text{'}}+{\left(x^2\right)}^{\text{'}}$$=-\frac{1}{x^2}-\frac{1}{2x\sqrt{x}}+2x$

Khi đó $g^{\text{'}}\left(1\right)=-\frac{1}{1^2}-\frac{1}{2.1\sqrt{1}}+2\cdot 1=\frac{1}{2}$

Do đó f'(0) – g'(1) = $\frac{1}{2}-\frac{1}{2}=0$ . Vậy f'(0) – g'(1) = 0.