y=2x√x, x0=1/3
[m]y=f(x_0)+f'(x_0)(x-x_0)[/m]
[m]f(x_0)=f(\frac{1}{3})=2*\frac{1}{3}\sqrt{\frac{1}{3}}=\frac{2}{3\sqrt{3}}=\frac{2\sqrt{3}}{9}[/m]
[m]f'(x)=(2x\sqrt{x})'=2\sqrt{x}+2x*\frac{1}{2\sqrt{x}}=2\sqrt{x}+\frac{x}{\sqrt{x}}=\frac{2x+x}{\sqrt{x}}=\frac{3x}{\sqrt{x}}=\frac{3x\sqrt{x}}{x}=3\sqrt{x}[/m]
[m]f'(x_0)=f'(\frac{1}{3})=3\sqrt{\frac{1}{3}}=\frac{3}{\sqrt{3}}=\frac{3\sqrt{3}}{3}=\sqrt{3}[/m]
[m]y=\frac{2\sqrt{3}}{9}+\sqrt{3}(x-\frac{1}{3})=\frac{2\sqrt{3}}{9}+x\sqrt{3}-\frac{\sqrt{3}}{3}=x\sqrt{3}+\frac{2\sqrt{3}-3\sqrt{3}}{9}=x\sqrt{3}-\frac{\sqrt{3}}{9}[/m]