# User Portlet Melvyn Stancliffe
Discussions
to calculate the derivative of a function at a given point example f[x_]:= 2x^2 +5x -Sin[x] f'[x]/.x->2 or f' pay attention to the definition of the function in WA . simply type "derivative of " (your function)...
example data = RandomInteger[{1, 100}, {20}]; {data, Map[Mod[#, 3] &, data]} or even {data,Mod[data,3]}
Not sure if its the optimum method but ... f[n_] := Table[Mean[RandomSample[data, 7]], {n}]; ListPlot[f]
ContourPlot[y^3 + 2 == Sqrt[x^2 + x + 1], {x, -10, 10}, {y, -5, 5}, Frame -> False, Axes -> True, AspectRatio -> Automatic] or ContourPlot[y^3 + 2 == Sqrt[x^2 + x + 1], {y, -10, 10}, {x, -5, 5}, Frame -> False, Axes -> True,...
domain function http://www.wolframalpha.com/input/?i=domain+%28Sqrt%5BLog%5B%285+x+-+14%29%2F%28x%5E2+-+3+x+%2B+2%29%5D%5D+%2B+Log%5B3+-+x%5D%29 range function ...
I think what is meant is how to compute the 6th derivative evaluated at x=1 D[f[x], {x, 6}] /. x -> 1
not sure if there is a shorthand for the unit step function but this works for me. Having checked , I think the shorthand is "theta" . laplace transform (t-1) UnitStep (t-1) ...
The upgrade was unnecessary (at least for this question) as I am using version 7. With version 10 , you dont need to use the Needs ["VectorAnalysis"] as this functionality is already included DotProduct[{3,2,4},{3,2,4},Cylindrical]//N ...
its Mathematica's way of saying c = pi/4 + n pi
f[x_]:=3x-5;g[x_]:=2-x^2; Composition[f,g][x]