# Is the derivative of a function the secant line?

I am just learning derivatives and I found the derivative of $4x-x^2$ to be $4-2x$. At point $(1,3)$ the tangent line is $2x+1$. Now when I graph this, the derivative $4-2x$ cuts through the function $4x-x^2$. Does that mean the derivative is the secant line?

The answer generally speaking is no. The derivative of a function at a point always represents the slope of the tangent line at that point. Sometimes the tangent line (by coincidence) crosses the curve somewhere else technically making it also a secant line, but there is no special meaning in this.

Based on your comment I misunderstood your qustion. The answer is still no in general. Try plotting the derivative of y=x^3.

• The moral is that the derivative describes local behavior near the point its evaluated at. Jul 17, 2013 at 3:18
• When I graph all three equations, the tangent line does look like a tangent line, and the derivative looks like a secant line. I did this for other problems and I get the same types of graphs: the derivative crosses the original function. Any help? Jul 17, 2013 at 3:22
• @Juan: Just a guess. What does the graph of your function look like (it should be a parabola that goes through (0, 0) and (4, 0) and it should open downward). Jul 17, 2013 at 3:32
• It represents the slopes at the different points. If you are looking at a curve $y=f'(x)$ then the y coordinate at a given value of $x$ is the slope of the tangent line (of the original function) at that value of $x$. Graphing your derivative tells us for instance that the slope of the tangents is positive if $x < 2$, zero if $x=2$, and negative if $x>2$. This can tell us a lot of useful things about the original function, for instance a negative slope means the function is decreasing and a positive slope means the function is increasing. Jul 17, 2013 at 5:23
• The derivative is a function which tells you the slope if you tell it what $x$ is. You will notice that a lot of us have been saying "the derivative at a point", this is because the value of the derivative depends on what $x$ is. If I give the derivative a value of $x$ then it tells me the slope of the tangent line at that $x$-coordinate on the graph of the original function. Jul 17, 2013 at 5:30

There is a strange misunderstanding that the graph of $f'$ should be related to the graph of a tangent line of $f$. Even more when the graph of $f'$ is itself a line.

But given any number $a$, the value of $f'(a)$ keeps track of the magnitude of a slope, namely, it is the value we assign to $m$ in $$\tau(x)=m(x-a)+f(a)$$ so that $$\frac{f(x)-\tau(x)}{x-a}\to 0 \text{ when } x\to a$$

Without appealing technicalities, this means $f'(a)$ is the slope of the line that best approximates $(\to 0)\;\;$ $f$ in the vicinity of $a\;$ ($x\to a$). When we plot $f'$, we're seeing how the value of the slope at $x$ varies.

ADD As an example, the derivative of $3x+x^3$ is a second degree polynomial $3+3x^2$. This is not a line at all: it simply tells us that the slope of $3x+x^3$ is always positive, and that it gets steeper and steeper for both large positive values and large negative values of $x$. It also says $3x+x^3$ has no horizontal tangents, since the derivative is never zero.