CODEWITHSUNDEEP
wolfram-mathematica
Koba
Koba

Reputation: 1544

Evaluating derivatives of functions of three variables in Mathematica

I am trying to evaluate the derivative of a function at a point (3,5,1) in Mathematica. So, thats my input:

   In[120]:= D[Sqrt[(z + x)/(y - 1)] - z^2, x]
   Out[121]= 1/(2 (-1 + y) Sqrt[(x + z)/(-1 + y)])
   In[122]:= f[x_, y_, z_] := %
   In[123]:= x = 3
             y = 5
             z = 1
             f[x, y, z]
   Out[124]= (1/8)[3, 5, 1]

As you can see I am getting some weird output. Any hints on evaluating that derivative at (3,5,1) please?

Upvotes: 1

Views: 1414

Answers (1)

rcollyer
rcollyer

Reputation: 10695

The result you get for Out[124] leads me to believe that f was not cleared of a previous definition. In particular, it appears to have what is known as an OwnValue which is set by an expression of the form 

f = 1/8

(Note the lack of a colon.) You can verify this by executing

g = 5;
OwnValues[g]

which returns

 {HoldPattern[g] :> 5}

Unfortunately, OwnValues supersede any other definition, like a function definition (known as a DownValue or, its variant, an UpValue). So, defining 

g[x_] := x^2

would cause g[5] to evaluate to 5[5]; clearly not what you want. So, Clear any symbols you intend to use as functions prior to their definition. That said, your definition of f will still run into problems.

At issue, is your use of SetDelayed (:=) when defining f. This prevents the right hand side of the assignment from taking on a value until f is executed later. For example,

D[x^2 + x y, x]
f[x_, y_] := %

x = 5
y = 6
f[x, y]

returns 6, instead. This occurs because 6 was last result generated, and f is effectively a synonym of %. There are two ways around this, either use Set (=)

Clear[f, x, y]
D[x^2 + x y, x];
f[x_, y_] = %

f[5, 6]

which returns 16, as expected, or ensure that % is replaced by its value before SetDelayed gets its hands on it, 

Clear[f, x, y]
D[x^2 + x y, x];
f[x_, y_] := Evaluate[%]

Upvotes: 6

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