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The first derivative tells me how the slope of the tangent line of each point to the graph varies, while the second derivative tells me how it varies as a true variation?

For example, having $f(x) = x^2$, $f'(x)= 2x$ (therefore the slope of each point of the line tangent to the point is equal to $2x$, for example at the point $x = 1$ is equal to $2$).

While the second derivative tells me how this variation varies, $f ''(x) = 2$. So $f''$ what are you telling me? I don't understand why it's a constant if the slope changes point to point.

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Consider $f(x)=2x$. It is a straight line with constant slope $2$, and that's what $f'(x)=2$ tells us.

Now consider $f(x)=x^2$. $f'(x)=2x$ tells us that the slope of $f$ is changing at each point, and that change is behaving like the function (straight line) $2x$. Then $f''(x)=2$ is telling us that the variation of the change of $f'$ is constant, which means that the slope of $f$ is changing, but it changes at a constant rate.

And yes, $f''$ does tell you how the slope of $f$ is varying, since the slope is given by $f'$ and the first derivative of $f'$ is $f''$. In an attempt to provide some intuition I'll use the following analogy:

Imagine $f$ as something that given a point $x$ gives you a position for $x$ on the graph. Then $f'$ is telling you how fast or slow a position is changing into the following positions (if $f'$ is positive it is changing upwards, and if $f'$ is negative it is changing downwards). So $f'$ could be though as "the speed" of the position moving. Then $f''$ is telling you how that change varies. If $f''$ is positive and big then $f'$ ("the speed") is going to get bigger rapidly, which means the following positions are going to change faster ($f''$ works as an "acceleration"). If $f''$ is negative then your $f'$ is going to get slower, so the position will change more and more slowly.

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  • $\begingroup$ So you are telling me that the slope varies in a ratio of 2? $\endgroup$ – user817101 Oct 25 '20 at 16:16
  • $\begingroup$ What do you mean with "varying in a ratio"? $\endgroup$ – Darsen Oct 25 '20 at 16:28
  • $\begingroup$ why the costant term is 2? $\endgroup$ – user817101 Oct 25 '20 at 16:29
  • $\begingroup$ For $f(x)=x^2$ you have $f'(x)=2x$. Now, for every unit you move (I mean going from $x$ to $x+1$) you get $f'(x+1)=2(x+1)=2x+2=f'(x)+2$, so the slope of the function is bigger by two units. This is what $f''(x)=2$ was saying. Just think that $f'$ is another function, so $f''$ gives you the slope of $f'$. It's like saying $f''$ is the slope of the slope. $\endgroup$ – Darsen Oct 25 '20 at 16:39
  • $\begingroup$ And the slope of slope what are you telling me? $\endgroup$ – user817101 Oct 25 '20 at 16:46

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