# tangent line, equation

Can you prove which is the tangent equation using derivative.

Why the tangent line $l$ through $(x_{0}, f(x_{0}))$ with slope $f'(x_{0})$ is: $$l(x)=f(x_{0})+f'(x_{0})(x-x_{0}) \mbox{?}$$

Thank!

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The equation of any line passing through $(x_{0}, f(x_{0}))$ is $\frac{y-f(x_0)}{x-x_0}=m$ where $m$ is the slope.

Now if the line is a tangent of $y=f(x)$ at $(x_{0}, f(x_{0}))$, the slope of the line = the slope of $y=f(x)$ at $(x_{0}, f(x_{0}))$ which is $f'(x_{0})$ .

So, the equation of the line which is a tangent of $y=f(x)$ at $(x_{0}, f(x_{0}))$ will be $\frac{y-f(x_0)}{x-x_0}=f'(x_0)$

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Ok :) now, how can I do the transition to the tangent plan. $\displaystyle z=f(x_{0},y_{0})+[\frac{\partial f}{\partial x}(x_{0},y_{0})](x-x_{0})+[\frac{\partial f}{\partial y}(x_{0},y_{0})](y-y_{0})$ –  Diana Lazar Aug 25 '12 at 7:50
Let $t = (x,y).$ Then, $z = f(t_0)+\nabla(t_0)(t-t_0)$ which is the $\mathbb{R}^2$ generalization of the $\mathbb{R}$ example first considered. –  Jeff Aug 25 '12 at 9:30

Hint : What is the equation of a line through a point with a given slope?

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