 The derivative of F at x0 •• At the beginning the function x → x³+1 is calculated.

```function F(x) {
with(Math) {
return  pow(x,3)+1
}}```  For x1 = 4, x2 = 4.00000001 I have 48: the derivative of x → x³+1 in 4.

To change F open the Code and modify F(x)

Save by clicking here (with the right mouse button) the file on your computer
(change the name if you want, but leave ".htm").
At this point you can use it to study any function.
Just open the source code, edit F(x), and run "save".
You can repeat this several times.

•• If I want the derivative in 0 of:

```function F(x) {
with(Math) {
if (x==0) {y = -1} else {y = (x*x+x)/(x*(abs(x)-1)) }
return y
}}
``` in 0, "from the left" the derivative in 0, "from the right" it is −2: •• If I want the derivative in 3 of:

```function F(x) {
with(Math) {
y = cos(x)/(2*pow(x,2)+3)
return y
}}
``` ```x1 = 2.999999999, x2 = 3.000000001 -> DF/Dx = 0.020218571670441426
x1 = 2.99999999, x2 = 3.00000001 -> DF/Dx = 0.020218571037598457
x1 = 2.9999999, x2 = 3.0000001 -> DF/Dx = 0.020218570947809964
x1 = 2.999999, x2 = 3.000001 -> DF/Dx = 0.020218570953527932
x1 = 2.99999, x2 = 3.00001 -> DF/Dx = 0.020218570951017416
x1 = 2.9999, x2 = 3.0001 -> DF/Dx = 0.020218570849868743
x1 = 2.999, x2 = 3.001 -> DF/Dx = 0.020218560713802937
x1 = 2.99, x2 = 3.01 -> DF/Dx = 0.02021754710226442
x = 3 -> F(x) = -0.04714249983811645```

I understand that, from a certain point on, rounding errors become predominant. I can round the derivative in 3 with 0.02021857095. If I did the calculation of the derivative formally I would get -((2*x^2+3)*sin(x)+4*x*cos(x))/(2*x^2+3)^2, that when x = 3 holds (rounding) 0.0202185709522633 (but for practical purposes 0.0202186 is more than enough).

•• If I want the derivative in 2 of:

```function F(x) {
with(Math) {
y = sqrt(x+sqrt(x+sqrt(x)))
return  y
}}``` ```x1 = 1.999999999, x2 = 2.000000001 -> DF/Dx = 0.3482589483968381
x1 = 1.99999999, x2 = 2.00000001 -> DF/Dx = 0.34825893491952525
x1 = 1.9999999, x2 = 2.0000001 -> DF/Dx = 0.34825892410451564
x1 = 1.999999, x2 = 2.000001 -> DF/Dx = 0.3482589245773269
x1 = 1.99999, x2 = 2.00001 -> DF/Dx = 0.3482589246294687
x1 = 1.9999, x2 = 2.0001 -> DF/Dx = 0.3482589247335425
x1 = 1.999, x2 = 2.001 -> DF/Dx = 0.3482589354979854
x1 = 1.99, x2 = 2.01 -> DF/Dx = 0.3482600119583381
x = 2 -> F(x) = 1.9615705608064609```

I understand that, from a certain point on, rounding errors become predominant. I can round the derivative in 2 with 0.3482589246. If I did the calculation of the derivative formally I would get (rounding) 0.34825892462483 (but for practical purposes 0.348259 is more than enough).

•• If I want the derivative in π/2 of:

```function F(x) {
with(Math) {
y = sin(x)*sin(x)/sin(x*x)
return  y
}}``` ```x1 = 1.5707963267948966, x2 = 1.57079632679 -> DF/Dx = 6.297569381461999
x1 = 1.5707963267948966, x2 = 1.5707963267 -> DF/Dx = 6.297662947849201
x1 = 1.5707963267948966, x2 = 1.570796326 -> DF/Dx = 6.29766367514587
x1 = 1.5707963267948966, x2 = 1.57079632 -> DF/Dx = 6.297662721957301
x1 = 1.5707963267948966, x2 = 1.5707963 -> DF/Dx = 6.297662147590565
x1 = 1.5707963267948966, x2 = 1.570796 -> DF/Dx = 6.297652236263873
x1 = 1.5707963267948966, x2 = 1.57079 -> DF/Dx = 6.297453845904398
x1 = 1.5707963267948966, x2 = 1.5707 -> DF/Dx = 6.294479377944526
x1 = 1.5707963267948966, x2 = 1.57 -> DF/Dx = 6.271434713300255
x = 1.5707963267948966 -> F(x) = 1.6018813708668798```

I understand that, from a certain point on, rounding errors become predominant. I can round the derivative in π/2 with 6.2976623. If I did the calculation of the derivative formally I would get (rounding) 6.2976630437795 (but for practical purposes 6.2977 is more than enough).

We can draw the graphs with JavaScript; for example, see sinsinsin  (see the code;  se also here).

Graphs can be plotted also with R or Desmos.