I looked for various answers and couldn't find anything helpful.What I know is the derivative of a curve at a point is the slope of tangent line drawn to the point.Nowwhat does second derivative mean slope of the slope of line?I know it in terms of velocity,acceleration etc. but it is hard for me to get it in math.

For example: The first derivative of $x^3$ is $3x^2$ but why is it not a straight line?In my book,it is written as slope of tangent.Please help regarding all these doubts in simple terms.I'm just in high school

  • $\begingroup$ See math.stackexchange.com/questions/329631/… $\endgroup$
    – John_dydx
    Jan 28, 2017 at 16:56
  • 1
    $\begingroup$ You can view second derivative as a measurement of the rate of first derivative changes. $\endgroup$
    – Fuhsuan Ho
    Jan 28, 2017 at 16:56
  • $\begingroup$ Your second question: at point $x$, the slope of tangent is exactly $3x^2$ for the function $f(x)=x^3$. This is exactly the meaning of first derivative. $\endgroup$
    – Fuhsuan Ho
    Jan 28, 2017 at 16:58

3 Answers 3


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.

Illustration of concavity of a graph on 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).


You are correct that the first derivative of $x^3$ is, $\frac{d}{dx}x^3=3x^2$. This is of course not a straight line, but rather a parabola. However, consider wanting to find the slope at a specific point along $x^3$, say $x=5$. The slope at this line is obtained simply by plugging $x=5$ into the derivative of $x^3$. That is, the slope, $m$, of $x^3$ is $3(5)^2=75$. This, I think you will agree is a straight tangent line. That is, at the point $x=5$ $y$ changes by $75$ as $x$ changes by $1$. In this sense, the derivative is not a tangent line, but rather a function to generate the tangent line at each point along a curve.

Similarly, one can think of the second derivative as the function which generates the rate of change of the first derivative at every point along the function. As stated above, if the second derivative is positive, it implies that the derivative, or slope is increasing, while if it is negative, implies that the slope is decreasing.

As a graphical example, consider the graph, $y=(x)(x-2)(x-3)$ which looks like this.

Graph of x(x-2)(x-3)

The derivative of this function is $3x^2-10x+6$, which we graph as the following.

Graph of 3x^2-10x+6

We can then read off the slope at any point along the original curve. For example, at $x=2$, we see that the slope is $3(2)^2-10(2)+6=-2$, which is negative. Looking at the graph of $x(x-2)(x-3)$, we can indeed see that $m$ for the tangent line drawn at $x=2$ would indeed have a negative slope.

As for the second derivative; it is given by, $6x-10$ (which again you may verify for yourself), and it looks like this.

enter image description here

The point at which it changes from positive to negative is at $6x-10=0\rightarrow x=\frac{10}{6}=\frac{5}{3}$. This does not mean that the slope is positive at $\frac{5}{3}$, rather, it means that the slope, on the interval $[\frac{5}{3},\infty)$ is increasing. On this interval the slope, generated by the first derivative, $3x^2-10x+6$ is increasing, such that $3(x+\Delta x)^2-10(x+\Delta x)+6>3x^2-10x+6$, for all $x$ in the interval, where $\Delta x$ is some small positive change in $x$. For example, the slope at $x=2$ which is in the interval is given by $-2$, whereas the slope of $x=3$, with $3$ obviously greater than $2$, is $3(3)^2-10(3)+6=3$, which is greater than the slope at $x=2$.

It's harder to see in this graph, but the significance of the second derivative is that, if you drew tangent lines along every point on the initial curve, that the slope of these tangent lines would be decreasing as you increased $x$ when negative, and increasing where positive.


First derivative of a curve is slope of tangent.

And $\frac{-1}{\text{slope of tangent}}$ is a slope of normal.

I think second derivative is used to show the rate of change of slope.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .