Sorry if I'm being too specific and for not showing an example but If you had a derivative of a straight line would the slope of the tangent line be the same as the straight line?
[Math] Is the derivative of a straight line the same of a tangent line
calculus
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It's not so simple to define what is a tangent straight line to a curve. Your definition ''the line that touch at one point'' can work only if the curve is a conic section ( so it is represented by a second degree equation), but also for a simple curve as $y=x^3$ you can see that all the straight lines that are intuitively tangent at some poin, meet the curve at another point, and the only line that ''touch at one point'' is the line $y=0$ that really is a tangent, but through the curve just at the point of tangency.
And you can also easily find functions that has the same line as an ''intuitive'' tangent at more than one point (as a simple exemple: $y=\sin x$ with the lines $y=\pm1$).
So a good definition of tangent can be done only at the some time as the definition of derivative, using a limit. In words:
The tangent line is the limit position of a secant line when the two points of intersection coincide.
Developing this intuition we come to the classical definition of the slope of the tangent as: $$ m=\lim_{h\rightarrow 0}\dfrac{f(x+h)-f(x)}{h}=f'(x) $$
Now, if the curve is a stright line, this definition can also be used and, yes, the tangent to a stright line is the same stright line.
The second derivative tells you something about how the graph curves on an interval.
If the second derivative is always positive on an interval $(a,b)$ then any chord connecting two points of the graph on that interval will lie above the graph. If the second derivative is always negative on the interval $(a,b)$ then any chord connecting two points of the graph on that interval will lie below the graph.
In the graph below of $y=x(x-1)(x+1)$ the graph has a negative second derivative on the interval $(-\infty,0)$ and a positive second derivative on the interval $(0,\infty)$ so it is concave down and concave up, respectively on the two intervals.
Another way of expressing the same idea is that if a continuous second differentiable function has a positive second derivative at point $(x_0,y_0)$ then on some neighborhood of $(x_0,y_0)$ the tangent line at $(x_0,y_0)$ lies below the graph (except at the point of tangency). If the second derivative is negative at the point of tangency the tangent line lies above the graph on some neighborhood of the point of tangency (except at the point of tangency).
Best Answer
recall the formal (limit) definition of the derivative:
$$f'(x) = \lim_{h \rightarrow 0} \dfrac{f(x + h) - f(x)}{h}$$
let $f(x) = ax + b$ where $a,b$ are real numbers. Then
$\begin{equation*} \begin{split} f'(x) &= \lim_{h \rightarrow 0} \dfrac{\big(a(x + h) + b\big) - (ax + b)}{h} \\ &= \lim_{h \rightarrow 0} \dfrac{ax + ah + b - ax - b}{h} \\ &= \lim_{h \rightarrow 0} \dfrac{ah}{h} \\ &= a \end{split} \end{equation*}$
which is a horizontal line (generally speaking the derivative at a specific point is the slope of the tangent line at that point) There are a couple of important things to note:
1) a line that is already horizontal will have a slope of 0 (that is $a$ = 0) so its derivative will always be 0
2) the derivative is a function of $x$ (our independent variable) so a vertical line does not have a derivative