Evaluate

Basic Examples

Take the numerical derivative of an exponential:

○
NDerivative
[Exp[x],x,1]
In[]:=
2.71828
Out[]=

Take the numerical derivative of a trig power:

○
NDerivative
Cos[x]
3
,{x,2},0
In[]:=
-3.
Out[]=

Scope

In[]:=

The expression and evaluation point may be complex-valued:

○
NDerivative
[Sin[x],x,πI]
In[]:=
11.592+1.32527×10
-10

Out[]=
○
NDerivative
[Cos[Ix],x,1+I]
In[]:=
0.634964+1.29846
Out[]=

Generalizations & Extensions

In[]:=

NDerivative
threads over its first argument:

○
NDerivative
[{Exp[x],Sin[x]},x,1]
In[]:=
{2.71828,0.540302}
Out[]=

Options

In[]:=

Method

In[]:=

Use the default
Method
"EulerSum"
when
expr
is not analytic in the neighborhood of
x
0
:

○
NDerivative
[Re[Cos[Iy]],y,1]
In[]:=
1.1752
Out[]=

Check:

D[ComplexExpand[Re[Cos[Iy]]],y]/.y->1//N
In[]:=
1.1752
Out[]=

An incorrect answer is obtained with
Method

NIntegrate
:

○
NDerivative
[Re[Cos[Iy]],y,1,Method->NIntegrate]
In[]:=
0.587601+8.39313×10
-17

Out[]=

Here is a derivative where the default method works poorly:

○
NDerivative
Exp[x
2
],{x,3},1
In[]:=
68.5073
Out[]=

The correct answer is:

DExp[x
2
],{x,3}/.x->1.
In[]:=
54.3656
Out[]=

In this case the expression is analytic, so
Method

NIntegrate
will work well:

○
NDerivative
Exp[x
2
],{x,3},1,Method->NIntegrate
In[]:=
54.3656+2.82473×10
-14

Out[]=

Scale

In[]:=

Use
Scales
to capture the region of variation:

○
NDerivative
[Sin[100x],x,0]
In[]:=
215.879
Out[]=

The scale of variation is around
.01
:

○
NDerivative
[Sin[100x],x,0,"Scale"->.01]
100.
Out[]=

A value of
Scales
that is too large can be compensated by increasing the number of terms:

○
NDerivative
[Sin[100x],x,0,"Terms"->11]
In[]:=
100.
Out[]=

Use
Scale
to specify directional derivatives. The left and right derivatives of the nonanalytic function
x
:

○
NDerivative
[Abs[x],{x,1},0]
In[]:=
1.
Out[]=
○
NDerivative
[Abs[x],{x,1},0,"Scale"->-1]
-1.
Out[]=

Check:

Plot[Abs[x],{x,-1,1}]
In[]:=

Complex directions may also be specified:

Check:

Check:

Terms

Increasing the number of terms may improve accuracy. Here is a somewhat inaccurate approximation:

Check:

Increasing the number of terms produces a more accurate answer:

Increasing the number of terms further can produce nonsense due to numerical instability:

Combining an increase in the number of terms with a higher working precision often will reduce the error:

WorkingPrecision

In[]:=

Using a higher working precision and additional terms produces an accurate answer:

Higher-order derivatives will again experience numerical instability:

An alternative is to increase the radius of the contour of integration:

Applications

In[]:=

Check:

Properties & Relations

In[]:=

The Wolfram Language has built-in code to compute derivatives of numerical functions:

The built-in numerical derivative code can be used. However, it is unable to capture the rapid oscillations:

The correct answer:

Check:

Neat Examples

In[]:=

Some fractional and complex derivatives can be computed: