# What is the actual meaning of second derivative?

I am confused why we use second derivative to find the maxima and minima. I cannot understand what is the meaning of second derivative. Also i have come across some formulae that is

• if second derivative is greater than zero then it is minima.
• if second derivative is less than zero then it is maxima

• if it is equal to zero then go on to higher order derivative.

Can anyone explain me what is the reason behind this formulae?

• refer this – Thulashitharan D Jul 16 at 11:24
• Imagine doing indoor skydiving in a vertical wind tunnel. Close your eyes. You want to know when you have reached (local) maxima or minima in height. What can you do? Well, the air blowing around you gives you a sense of speed and in your stomach (and wherever else) you can sense the acceleration you are experiencing. Putting both together, you are at local maximum when you experience no vertical speed, but feel a downward drag, and a local minimum when you experience no vertical speed, but feel an upward push, right? – k.stm Jul 16 at 11:40
• By the way, it’s kind of strange how a non-effort non-specific, but very typical question by a new user about an extremely common and elementary beginner’s topic for which a plethora of information and explanations is available online (without anything being referenced in the question) gathers attention from four or five high or very high reputation users with no upvote on any answer and two downvotes on the question itself … – k.stm Jul 16 at 11:46
• @k.stm i have searched on the internet and found some fomulae that is- *if second derivative is greater than zero than it is minima. * is second derivative is less than zero than it is maxima and if it is equal to zero than go on to higher order derivative. I just wanted the reason behind this formulae thats why i have asked this question. I really admit that it is my mistake that i should be more specific. Next time i will surely asked with full of information regarding my question. – Anwesh Panda Jul 16 at 12:05
• @k.stm The question isn't a homework question and it's not a question for which one could "show work." So it's not really violating usual etiquette. So no downvotes. Upvotes, if any, will come when everyone finally gets out of bed. – B. Goddard Jul 16 at 12:49

The extrema are found where the derivative is zero. As zero has no sign, you can't tell a minimum from a maximum.

A minimum is where the slope goes from negative to positive, hence the first derivative is decreasing and conversely a maximum is where the slope goes from positive to negative, hence the first derivative is increasing.

So the sign of the second derivative allows you to tell la minimum from a maximum.

If the second derivative is zero, you need more criteria.

One intuitive example is the height of a moving object moving in one dimension.

Its first derivative is velocity, if the change in distance is smooth enough. Turning points in distance (i.e. local maxima or minima) happen when the velocity is zero, but knowing the velocity is zero does not tell you whether the height is a maximum or a minimum.

So the next step is to look at the second derivative of height, which is acceleration. If the velocity is zero and the acceleration is negative (i.e. downwards) then you can conclude the velocity is changing from positive (upwards) to negative (downwards) that height is at a maximum. Similarly if velocity is zero and the acceleration is positive (upwards), then you can conclude that height is at a minimum.

You can think of the 2nd derivative as what is the rate of change of the rate of change or how curved does it look. look here for a good info https://www.youtube.com/watch?v=BLkz5LGWihw&list=PLZHQObOWTQDMsr9K-rj53DwVRMYO3t5Yr&index=10

• Thanks very much – Anwesh Panda Jul 16 at 13:23

If you're looking at the graph of the function, the first derivative is the slope of the tangent line. Derivatives tell you how something is changing. If the second derivative is positive, it means the first derivative is increasing.

Imagine the tangent line on a curve at a point while the point moves from left to right. If the slope is increasing, then the tangent line is rotating counter-clockwise. If the slope is decreasing, then the tangent line is rotating clockwise. So you have this rule: Second derivative positive means counter-clockwise rotation. Second derivative negative means clockwise rotation.

Now further imagine what these rotations mean about the shape of the curve. If the rotation is counter-clockwise, the curve must be concave up. If the curve is concave up, and you happen to be at a critical point, then that critical point must be a minimum. (The first derivative is $$0$$ here, and since the slope is increasing, it has to be negative on the left and positive on the right.)

In sum, second derivative positive means tangent line is rotating counter-clockwise. In turn, this means the curve is concave up. In turn, this means a critical point is a minimum.

• You have written this statement"If the second derivative is positive, it means the first derivative is increasing." can you explain me how? – Anwesh Panda Jul 16 at 12:10
• Derivatives are rates of change. If the derivative of something is positive, the rate is increasing. The derivative of the derivative being positive means the derivative in increasing. – B. Goddard Jul 16 at 12:47
• If second order derivative imply the rate of change of slope than what will 3rd order derivative imply? – Anwesh Panda Jul 16 at 13:01
• If the second derivative is large, the tangent line is rotating quickly, which means the curve is curvier. The third derivative measures how fast or slow the curve is getting curvier. – B. Goddard Jul 16 at 13:06
• Ok thanks i understood it. – Anwesh Panda Jul 16 at 13:07

The second derivative measures the variations of the first. Therefore if $$f''(x)$$ has a constant sign on an interval $$I$$, say $$f''(x)>0$$, and $$f'(x_0)= 0$$ for some $$x_0\in I$$, it means that, in $$I$$ \begin{align} f'(x)<0,\enspace &\text{ hence f(x) is decreasing for }\; x0,\enspace &\text{ hence f(x) is increasing for }\; x>x_0 , \end{align} which shows $$f(x)$$ has a local minimum at $$x_0$$.

For similar reasons, if $$f''(x)<0$$ on $$I$$, $$f'(x)$$ decreases on $$I$$ and we conclude to a local maximum.