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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?

Pedro
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user5826
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2 Answers2

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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.

Spencer
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  • The moral is that the derivative describes local behavior near the point its evaluated at. – Spencer Jul 17 '13 at 03:18
  • 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? – user5826 Jul 17 '13 at 03:22
  • @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). – Owen Sizemore Jul 17 '13 at 03:32
  • @Juan, I think I misunderstood your original question and added a part to my answer. Does this answer your question? – Spencer Jul 17 '13 at 03:43
  • At $(2,4)$ the tangent is horizontal, so is not a secant. For a range around $x=2$ it will still not cut. You could solve the simultaneous equations to find where it starts cutting and being a secant. – Ross Millikan Jul 17 '13 at 03:45
  • The parabola has points (0,0), and (4,0, opens downward. the derivative 4-2x, passes through the parabola around (1,2) and again at around (5,7). – user5826 Jul 17 '13 at 04:14
  • @Juan: Your calculation seems wrong. (5, 7) is not a point on your equation, or the derivative or the tangent line at (1, 3). Also there is nothing wrong with the derivative function $f'$ looking like a secant line. The derivative $\textit{at a point}$ gives the slope of the tangent line through that point. Since $f'(x)=4-2x$ then $f'(1)=2$. Thus you get the slope of the tangent line is at (1, 3) is 2. – Owen Sizemore Jul 17 '13 at 04:24
  • I see. so the derivative of a function will always give the slope of the tangent line through that one point? – user5826 Jul 17 '13 at 04:39
  • Through any point you want. For instance if you want the slope of the tangent at $x=4$ evaluate $f'(4)$. If you want the slope of the tangent at $x=1000$ evaluate $f'(1000)$. – Spencer Jul 17 '13 at 04:42
  • There is nothing special about where the derivative crosses the original function. (at least nothing important for a calculus 1 course) – Spencer Jul 17 '13 at 04:42
  • i get it. So what is equation 4-2x supposed to represent? I got that when I solved lim h->0 (f(x+h)-f(x))/h for y=4x-x^2 – user5826 Jul 17 '13 at 05:13
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    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. – Spencer Jul 17 '13 at 05:23
  • You might also notice that the vertex of your parabola occurs at $x=2$ where the derivative is $0$ (crosses the x-axis). – Spencer Jul 17 '13 at 05:24
  • Is a "derivative" a slope (slope of the tangent line), or is it a function? – user5826 Jul 17 '13 at 05:25
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    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. – Spencer Jul 17 '13 at 05:30
  • Also I think it makes sense why the derivative has to be a function. Different parts of the graph clearly have different tangent lines so their slopes will be different. – Spencer Jul 17 '13 at 05:31
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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.

Pedro
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