DIFF returns the derivative in a location in a two- or more dimensional space
Arguments
In this example, the TeLiTab is addressed in the Dataslot. The function y is defined as y = DIFF(1, 2, "XC", "YC", x, 1) With the following Telitab set in the Data slot:
|DIFF1|
0
2
"XC" "YC"
"1" 1 1
"2" 2 4
"3" 3 9
"4" 4 16
"5" 5 25
"6" 6 36
"7" 7 49
"8" 8 64
"9" 9 81
"10" 10 100|
For x = 2.5, the function returns
y=5 NOTE: In case you apply the symbolic addressing of the columns for the description of the point on the curve or surface to compute the differential for, e.g. "Par_x" and "Par_y", please make sure that your Telitab set contains these names. If not, an error message is generated and the calculation is stopped.
In direct definition, the points of the curve are stated in the Relation itself. This method can only be used for 2D derivatives, the syntax is: DIFF(Pno%, Ndim%, "ColLab$_1,.., "ColLab$_Ndim%", Xint, [DirivNo%]) If Pno%=0 then all x_i and y_i values should be numeric expressions. The minimum number of x,y data points Ndim% in the list is 2 in which case the interpolation (and differentiation) is performed linear. Let the function y be defined as
y = DIFF(0, 4, 1, 1, 2, 4, 3, 9, 4, 16, x, 1) For x=2.5, this function returns
y=5
We have a relation:
DataSet2# = TEXTITEM$(1)
With in its dataslot:
TEXTITEM1= |0
3 "X" "Y" "Z"
"1" 1 1 2
"2" 1 4 8
"3" 1 9 18
"4" 1 16 32
"5" 1 25 50
"6" 1 36 72
"7" 1 49 98
"8" 1 64 128
"9" 1 81 162
"10" 1 100 200
"11" 2 1 12
"12" 2 4 18
"13" 2 9 118
"14" 2 16 132
"15" 2 25 150
"16" 2 36 172
"17" 2 49 198
"18" 2 64 1128
"19" 2 81 1162
"20" 2 100 1200
"21" 3 1 22
"22" 3 4 28
"23" 3 9 218
"24" 3 16 232
"25" 3 25 250
"26" 3 36 272
"27" 3 49 298
"28" 3 64 2128
"29" 3 81 2162
"30" 3 100 2200|
And use the following relation to determine the derivative:
Calculated_Value=DIFF(DataSet2#,3,"X","Y","Z", Input_Value_x, Input_Value_y, OptionalDirivNo)
With Input_Value_x = 1,Input_Value_y = 2 and OptionalDirivNo = 2 this gives Calculated_Value=2