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**666 bro****Member**- From: Flatland
- Registered: 2019-04-26
- Posts: 619

After complete revision of matrices and determinants now I'm feeling clear confidence in matrices and I've started revising calculus especially derivatives. My question to you is how would I understand the geometrical interpretation of an derivative ?

"An equation for me has no meaning, unless it expresses a thought of God"- Srinivasa ramanujan

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 8,931

hi 666 bro

I'm glad that your work on matrices has gone well.

The derivative of a function is the slope (gradient) of the graph at points along the curve.

example.

y = x^2

Just by looking at the graph we can see that at x = 0 the gradient is zero; that it rises up through positive values and has matching but negative values when x < 0

Differentiating the function we get dy/dx = 2x and this has the right gradient properties. When x = 0 2x is zero. At say x = 3 the gradient is 6 and at x = -3 it is -6.

There's lots on this on the MIF page https://www.mathsisfun.com/calculus/der … ction.html and there are links to other pages on calculus from there.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**666 bro****Member**- From: Flatland
- Registered: 2019-04-26
- Posts: 619

Now I've understood clearly geometrical interpretation of derivative now I'm learning maxima and minima of derivatives the only thing confuses me is that why it's required to do second derivative test for

finding maximum and minimum?

"An equation for me has no meaning, unless it expresses a thought of God"- Srinivasa ramanujan

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 8,931

hi 666bro,

It's great to hear that things are going well for you.

I'll illustrate what's happening with an example.

I've chosen the function

When we differentiate we get

and second derivative

Here are the graphs of (black) the function; (red) the first derivative; (green) the second derivative.

I've put the three graphs one after the other, lined up in the x direction but one on top of the other in the y direction. For this reason don't worry about the numbering on the axes; you don't need it. But the changes in gradient for the function are aligned correctly with the first and second derivatives.

The function is a cubic. When you differentiate it you get a quadratic. The quadratic has two zeros and these line up with the turning points on the cubic. When you differentiate again you are finding the gradient function of dy/dx, and you get a linear function.

On the (black) function graph you can see the two turning points; the first a maximum; the second a minimum; but let's say we don't know this yet. Just to the left of a maximum the function has positive gradient; just to the right it is negative. At the turning point the gradient is zero of course.

You can see this on the red graph. Just to the left of the function's turning point the red graph is above the axis so it's values are positive. The red graph crosses its axis at the turning point and after that it has negative values. In a similar way the next turning point (a minimum) has gradient negative to the left of the turning point; zero at the turning point; then positive.

So you can tell if a turning point is a maximum by looking to see if the first derivative goes from positive, through zero, to negative. And a minimum will have first derivative that goes from negative, through zero, to positive.

One way to investigate this is to draw the graphs; another is to calculate gradients just left and right of the turning point. But the second derivative gives a quick way to find out without needing to do any graph drawing. Here's how:

Function has a maximum. Gradient function goes from positive, through zero to negative. So its gradient function must be negative at that point as dy/dx has a reducing gradient.

Function has a minimum. Gradient function goes from negative, through zero, to positive. So its gradient function must be positive at that point as dy/dx has an increasing gradient.

In my example

so the turning points are at x = - √ (1/3) and + √ (1/3) At x = - √ (1/3) the second derivative is negative => this turning point is a minimum.At x = + √ (1/3) the second derivative is postive => this turning point is a maximum.

Note: I don't even have to do the full calculation here; I just need to know if the second derivative is positive or negative at each turning point; so it's a quick way to tell.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**666 bro****Member**- From: Flatland
- Registered: 2019-04-26
- Posts: 619

I've understood the concept of maximum and minimum, now I'm studying about first derivative test and i came to know that it's somehow relates to the concept of increasing and decreasing functions. My question is are these topics are same or different if it's

Different how they differ?

"An equation for me has no meaning, unless it expresses a thought of God"- Srinivasa ramanujan

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 8,931

hi 666bro,

I'll use the example from above.

We know there's a turning point at √(1/3) ≈ 0.57735

Investigate the gradient just left and just right of the turning point. I've chosen x = 0 and x = 1 to make the calculation easy.

The top line in the table shows the x values chosen and the second line shows the gradient at those points.

Below that I've sketched those gradients and the sketch shows it's a minimum.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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