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I am restudying Calculus on my own, and I am a little bit stuck on the definition of the slope of a tangent line to a point on a curve. I understand the definition somewhat, but I got to wondering about why the current definition was chosen. Basically, I do not understand why we do not take the following limit as the definition: $$\text{slope} = \lim\limits_{Q\to P}\text{slope}_{\text{sec}} = \lim\limits_{y_1\to y_0}\frac{y_1-y_0}{x_1-x_0}$$ Instead we use the following: $$\text{slope} = \lim\limits_{Q\to P}\text{slope}_{\text{sec}} = \lim\limits_{x_1\to x_0}\frac{y_1-y_0}{x_1-x_0}$$

From Simmons Book: pg. 53

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    $\begingroup$ How would you find the derivative of $f(x)=1$ then? The point is, if $f$ is a function, for any $x$, there is exactly one $y$ s.t. $f(x)=y$. $\endgroup$ – enedil Mar 7 at 0:47
  • $\begingroup$ @enedil Thanks for the comment. You have a good point. $\endgroup$ – SebastianLinde Mar 7 at 15:35
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Using a graph, you could do it the way you suggest relatively easily. However, more generally, if you are using a function definition instead, I believe the basic issue is that you would need to find the inverse of the function. As enedil commented, for $x_1 \to x_0$, you can determine $y_1$ for each $x_1$ easily using the function definition, i.e., $f(x_1) = y_1$. However, if you tried $y_1 \to y_0$ instead, to determine the corresponding values of $x_1$, you would need to use the function inverse, i.e., $x_1 = f^{-1}(y_1)$. However, not all functions have inverses and, even if they do, it requires more work to calculate them than necessary.

In summary, to have just one definition covering both using a graph and a function, it makes more sense to have $x_1 \to x_0$.

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  • $\begingroup$ Thanks for your answer. I get the point. $\endgroup$ – SebastianLinde Mar 7 at 15:34

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