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How do I prove that $$\frac{\Delta y}{\Delta x} = \frac{f(x+Δx)-f(x)} {Δx}.$$

I know that this is the slope formula to find the derivative of a function $y=f(x)$ and I know that the formula for a line tangent to $f$ in $(x_1, y_1)$ is: $$m=\frac{y-y_1}{x-x_1}$$ But really I do not know why we replace $y$ by $f(x+Δx)$ and $y1$ by $f(x)$.

Clear example with images would be appreciated.

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There is a simple explanation here (with colorful images as well): Not sure why you had to ask it here. – user26486 Jun 26 '14 at 12:24
up vote 7 down vote accepted

Let $$y=f(x).$$

Adding an icrement $\Delta x$ to argument results in an increase of $\Delta y$ in the function, or $$y+\Delta y=f(x+\Delta x)$$ $$\therefore \Delta y=f(x+\Delta x)-y$$ $$=f(x+\Delta x)-f(x).$$ Dividing both sides by $\Delta x$, we have $$\frac{\Delta y}{\Delta x}=\frac{f(x+\Delta x)-f(x)}{\Delta x}.$$ Edit. Alternatively say at a point $x_1$ a function has a value $f(x_1)$ and then at another point $x_2$ then the function has a value $f(x_2)$. The slope of the function between these two points is given by $$m=\frac{f(x_2)-f)(x_1)}{x_2-x_1}.$$ Now let us assume that the $x_2-x_1=\Delta x$, which means that the distance between the points $x_1$ and $x_2$ is $\Delta x$. From this we se that $x_2=\Delta x+x_1$, therefore $$m=\frac{f(x_1+\Delta x)-f(x_1)}{\Delta x}.$$ Dropping the subscript $1$ from $x_1$ means we can treat $x_1$ as arbitrary, thus $$m=\frac{f(x+\Delta x)-f(x)}{\Delta x}.$$ Now,$\Delta y$ is the increase in the function as the independent variable, $x$, increase by $\Delta x$. Thus $m=\frac{\Delta y}{\Delta x}$.

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Check out a diagram, here. – M. Knight Jun 26 '14 at 10:26
thank you so much, that was very useful for me, thanx for your efforts. – Mohammad Haidar Jun 26 '14 at 10:35

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