Why is the derivative graph of $y=x^2$ linear and not some sort of curve? I know that $\frac{d}{dx}x^2=2x$, but I am not talking computing algebraically, but more conceptually, thinking in terms of how the slope of tangent changes. When I look at the graph, why is the slope of $x^2$ changing at a linear rate?
Alternatively, why is the derivative of $y=ln(x)$ some kind of curve and not linear? I assume the rate of "curviness" of $x^2$ is different than $ln(x)$ which gets flat in a hurry.
(Does 2nd derivative and concavity explain this? Rate of change of the rate of change? )
My last example in the scan below summarizes my question. Why is it one of those, and not the other 3? All 4 start with a slope of 0, then 1, then back to 0. What is the connection, graphically ?
Thank you!
$\endgroup$ 23 Answers
$\begingroup$It's easy to visually see that the derivative of $x^2$ is $2x$, by not thinking in terms of how the slope changes, but plotting the slope of the tangents at points and seeing.
As to 'why' it is $2x$... It is not clear what kind of answer you are looking for. I intepret your wish as wanting a more ontological answer.
Think about two consecutive square numbers. Well $(x+1)^2 =x^2+2x+1$.
So when we add one to $x$, we add $2x+1$ to $f(x)$. If we add $2$ to $x$ we add $4x+4$ to $f(x)$. That is when we add some $c$ to $x$, we add $2cx + c^2$ to $f(x)$. This gives us a slope term of $2x+c$.
But for $x$ much larger than $c$ this is $~2x$.
Now, when we take the tangent on the $x^2$ curve, we are making $c$ as small as possible to get a more accurate rate of change at $x$ and this can (atleast in my head) be thought of as making the tangent line intersect the $x^2$ curve as much as possible. And assuming all lines have same width (euclidean assumption), we get the largest area of intersection when $c$ is smallest because this results in the part of the tangent line leaving the $x^2$ curve having the smallest area possible. That is because a line is defined as having constant width, and a shortest $c$ implies shortest possible length of the tangent line that has left the $x^2$ curve between $x$ and $x+c$. So the area of tangent that has left the $x^2$ curve in this zone is smallest. Whilst the rest of the tangent line that is not intersecting the $x^2$ has infinite area no matter what $c$ .... hence my interpretation of taking a tangent at $x$ on $x^2$ is defined as increasing the intersection between tangent line and curve as much as possible about $x$ (of course this wouldnt work for all curves, but it would for $x^2$.
Anyways, in pure mathematics we assume that the smallest $c$ is in fact small enough to make $2x+c = 2x$ ignoring other physical variables.
In some sense, the slope being $2x$ for $f(x)=x^2$ is a consequence of our chosen deconstructions which give us nice aesthetical answers.
$\endgroup$ 1 $\begingroup$I think you're looking for a "visual" aid to tell you whether or not the derivative of a curve will look linear. You have seen that parabolas have linear derivatives, but the converse is also true: if the derivative of a (reasonable) function is piecewise linear, then the function is piecewise quadratic.
In other words, the visualization skills to see that a smooth-looking graph is in fact going to have a linear derivative (the difference between $x^2$ and $\ln(x)$ that you are seeing) are the exact same as the visualization skills to see that a smooth-looking graph is parabolic. How much you can "see" this is a little subjective and will depend on the viewer.
$\endgroup$ 2 $\begingroup$What you're attempting to reason is why a derivative to a graph $f(x)$ is linear or non-linear. If you do a simple test visually say, the first segment of your reference graph is concave up and positive slope. The positive slope notifies that the graph of the derivative will be in the positive terminal. The concave up quality of the initial part of the graph assumes that the derivative is increasing. However, the question is at what rate is the rate of change increasing? This gives that the graph of the derivative is concave up because the original function is increasing.
$\endgroup$