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Multiple Derivative Task

I have a major problem in Mathematica that has puzzled me for a while. I want to take the mth derivative of x*Exp[t*x], and then evaluate this at x=0. But the following does not work correctly. Please share your thoughts.

D[x*Exp[t*x], {x, m}] /. x -> 0

Also what error means

General::ivar: 0 is not a valid variable.

Edit: my previous example ( D[Exp[t*x], {x, m}] /. x -> 0) was trivial. So I made it harder. :) My question is: how to get him to first evaluate the derivative, and then make a replacement.

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4 answers

, ( ) Mathematica , , , . , D[expr,{x,m}] , , x->0, , .

, , , m- , .

Derivative[m][Function[x,x Exp[t x]]][0]


Derivative[m][# Exp[t #]&][0]


In[2]:= Table[%, {m, 1, 10}]
Out[2]= {1, 2 t, 3 t^2, 4 t^3, 5 t^4, 6 t^5, 7 t^6, 8 t^7, 9 t^8, 10 t^9}

, m'th , SeriesCoefficient , :

In[3]:= SeriesCoefficient[x*Exp[t*x], {x, 0, m}]
Out[3]= Piecewise[{{t^(-1 + m)/(-1 + m)!, m >= 1}}, 0]

m- . PieceWise , .

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, mth- w.r.t. . t , .

Mathematica: D [Exp [t * x], {x, m}] = t ^ m Exp [t * x]

x , t ^ m, lim (Exp [t * x]) = 1. ?

:. x * exp (t * x)

m- w.r.t. x Wolfram Alpha:

t ^ (m-1) * exp (t * x) (t * x + m)

, x = 0, m * t ^ (m-1).

Q.E.D.

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, , . Mathematica ; x 0, , :

D[Exp[t*0],{0,m}]

, , , 0 .

+1

, :

:

D[Sin[x], {x, 1}]

x

Cos[x]

, x {x, 1} x Sin [x], , .
x , x, , . :

In[85]:= z=x^2;
D[Sin[z],{x,1}]
Out[86]= 2 x Cos[x^2]

? ! .

, , .

, , . . , , :

f[x_] := x*Exp[t*x];
g[y_, m_] := D[f[x], {x, m}] /. x -> y;
{g[p, 2], g[0, 1]}

:

{2 E^(p t) t + E^(p t) p t^2, 1}

!

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