Second Derivative

MathsIsFun · 2012-12-31 12:51:25

Just drafted this: Second Derivative

Comments/suggestions welcome!

bobbym · 2012-12-31 13:01:54

Hi MIF;

Works good and Happy New year!

Bob · 2012-12-31 22:50:26

Happy New Year MathsIsFun,

That page works nicely for me.

Do you have plans to extend it to show what f '' can be used for?

Bob

MathsIsFun · 2012-12-31 22:57:14

bob bundy wrote:

Do you have plans to extend it to show what f '' can be used for?

I do NOW ...

... can you help?

Jai Ganesh · 2013-01-01 01:08:08

Hi MathsIsFun,

Nice Page! Works well with me!!!

noelevans · 2013-01-01 07:56:52

It would be nice to have a link to a calculator that can calculate these derivatives for the common functions and graph f, f' and f'' in different colors on the same set of axes.

I think such programs exist, but are probably part of a larger program or are not free. A simple
free link to a program that just does the derivatives and graphing would be nice.

scientia · 2013-01-01 08:09:00

You mean this?

http://www.wolframalpha.com/input/?i=di … e+x^2&lk=4

noelevans · 2013-01-01 08:22:17

That's nice and a good site to remember. It only lacks a graph of f, f' and f'' on the same set of
axes.

bobbym · 2013-01-01 08:27:31

Hi noelevans;

Plot[{x^3,differentiate(x^3,x),differentiate(differentiate(x^3,x),x)},{x,-5,5}]

or
saving some typing

Plot[{x^3,D(x^3,x),D(D(x^3,x),x)},{x,-5,5}]

noelevans · 2013-01-01 08:51:33

Thanks bobbym,

I used copy and paste to put Plot[{x^3,D(x^3,x),D(D(x^3,x),x)},{x,-5,5}] into the input instead of
the differentiate command and it plotted all three. So that site does much more than just differentiate. Neat!

bobbym · 2013-01-01 09:13:31

Hi noelevans;

Your welcome!

Hi MIF;

Is it possible to get the equation for the tangent line of the crosshairs displayed?

MathsIsFun · 2013-01-01 10:35:59

bobbym wrote:

Is it possible to get the equation for the tangent line of the crosshairs displayed?

As in "y=3.105x+2.09" ?
Or perhaps just the slope.

bobbym · 2013-01-01 10:43:37

Hi;

It is great that I have a choice! The equation would be really nice.

Bob · 2013-01-01 10:46:11

bob bundy wrote:
Do you have plans to extend it to show what f '' can be used for?
I do NOW ...
... can you help?

Sure will.

Give me a few days.

Bob

Bob · 2013-01-13 23:20:25

hi MathsIsFun

Here's a first installment.

Example.

Find the turning points, and identify if each is a maximum, minimum or point of inflection.

The turning points of a function are places on the curve where the gradient is momentarily zero.

First differentiate the function.

Solving this quadratic gives

Substituting these into the original function means the turning points are at (1/3 , -0.18519) and at (-1 , 1)

The first graphs below show the function in blue and the derivative function in red.

Notice that the two places where the red curve crosses the x axis line up with the turning points of the function.

Now, if you have the graph in front of you, it is obvious that the first is a maximum and the second a minimum.

But what if you have to prove it without referring to the graph ?

Differentiate the gradient function.

So the gradient of the red curve at x = -1 is negative. Without having to see the graph I know the red curve crosses the x axis from positive values to negative values as x increases from less than -1 to more than -1.

So I know the blue curve goes from a positive gradient, through zero, to a negative gradient. So it must be a maximum.

So the gradient of the red curve at x = 1/3 is positive. Without having to see the graph I know the red curve crosses the x axis from negative values to positive values as x increases from less than 1/3 to more than 1/3.

So I know the blue curve goes from a negative gradient, through zero, to a positive gradient. So it must be a minimum.

The red curve is repeated along with its gradient function (the double differentiated function) in blue on the second graphs below. I have put little + and - signs close to the turning point values to show how the red curve goes from + to - or from - to +

Rule for identifying maximums and minimums:

Bob

MathsIsFun · 2013-01-14 08:32:13

Great example, will adapt it to web page.

Bob · 2013-01-14 08:40:26

Thank you.

I'm a bit dis-satisfied with the explanation of gradients of gradient graphs. It starts getting complicated at that stage. I've just had a thought. Maybe a question and answer stage before the final rule. I'll work on it tomorrow.

Would you like a piece on "points of inflexion" ?

Bob

Bob · 2013-01-15 11:58:47

hi MathsIsFun

Here is an alternative way of presenting the analysis of gradients as a series of questions. By making the reader answer step by step questions I think it leads them through the logic more carefully.

You could also add a close up picture of the relevant graph area.

I wrote:

Differentiate the gradient function.

So the gradient of the red curve at x = -1 is NEGATIVE.

Try to answer these questions:

Q1. Just to the left of x = -1 on the red graph the value of dy/dx is POSITIVE or NEGATIVE

Q2. Just to the right of x = -1 on the red graph the value of dy/dx is POSITIVE or NEGATIVE

Now think about the original function.

Q3. Just to the left of x = -1 on the blue graph the curve is sloping UPWARDS or DOWNWARDS.

Q4. Just to the right of x = -1 on the blue graph the curve is sloping UPWARDS or DOWNWARDS.

Q5. So the turning point at x = -1 is a MAXIMUM / MINIMUM.

So the gradient of the red curve at x = 1/3 is POSITIVE.

Try to answer these questions:

Q6. Just to the left of x = 1/3 on the red graph the value of dy/dx is POSITIVE or NEGATIVE

Q7. Just to the right of x = 1/3 on the red graph the value of dy/dx is POSITIVE or NEGATIVE

Now think about the original function.

Q8. Just to the left of x = 1/3 on the blue graph the curve is sloping UPWARDS or DOWNWARDS.

Q9. Just to the right of x = 1/3 on the blue graph the curve is sloping UPWARDS or DOWNWARDS.

Q10. So the turning point at x = 1/3 is a MAXIMUM / MINIMUM.

i wrote:

Rule for identifying maximums and minimums: .................................

finish as before

Bob

MathsIsFun · 2013-01-16 10:38:46

Hi Bob,

I will have a look at that and get back to you (when I have completed a few other things I am doing now)

Yours, Rod

Mathegocart · 2013-01-17 14:13:08

Great,but make it a it longer.great calc.

Math Is Fun Forum

#1 2012-12-31 12:51:25

Second Derivative

#2 2012-12-31 13:01:54

Re: Second Derivative

#3 2012-12-31 22:50:26

Re: Second Derivative

#4 2012-12-31 22:57:14

Re: Second Derivative

#5 2013-01-01 01:08:08

Re: Second Derivative

#6 2013-01-01 07:56:52

Re: Second Derivative

#7 2013-01-01 08:09:00

Re: Second Derivative

#8 2013-01-01 08:22:17

Re: Second Derivative

#9 2013-01-01 08:27:31

Re: Second Derivative

#10 2013-01-01 08:51:33

Re: Second Derivative

#11 2013-01-01 09:13:31

Re: Second Derivative

#12 2013-01-01 10:35:59

Re: Second Derivative

#13 2013-01-01 10:43:37

Re: Second Derivative

#14 2013-01-01 10:46:11

Re: Second Derivative

#15 2013-01-13 23:20:25

Re: Second Derivative

#16 2013-01-14 08:32:13

Re: Second Derivative

#17 2013-01-14 08:40:26

Re: Second Derivative

#18 2013-01-15 11:58:47

Re: Second Derivative

#19 2013-01-16 10:38:46

Re: Second Derivative

#20 2013-01-17 14:13:08

Re: Second Derivative

Board footer