I have the expression D[f[x, y], x]
, and I want to substitute f[x,y]
with x*y
, I tried the following:
D[f[x, y], x] /. {f[x,y] -> x*y}
and
D[f[x, y], x] /. {f -> x*y}
But neither worked. Would appreciate your help! Thanks.
I have the expression D[f[x, y], x]
, and I want to substitute f[x,y]
with x*y
, I tried the following:
D[f[x, y], x] /. {f[x,y] -> x*y}
and
D[f[x, y], x] /. {f -> x*y}
But neither worked. Would appreciate your help! Thanks.
The FullForm
of the derivative in your expression is
In[145]:= D[f[x,y],x]//FullForm
Out[145]//FullForm= Derivative[1,0][f][x,y]
This should explain why the first rule failed - there is no f[x,y]
in your expression any more. The second rule failed because Derivative
considers f
to be a function, while you substitute it by an expression. What you can do is:
In[146]:= D[f[x,y],x]/.f->(#1*#2&)
Out[146]= y
Note that the parentheses around a pure function are essential, to avoid precedence - related bugs.
Alternatively, you could define your r.h.s through patterns:
In[148]:=
fn[x_,y_]:=x*y;
D[f[x,y],x]/.f->fn
Out[149]= y
HTH
Nothing new, just the way I usually think of it:
D[f[x, y], x] /. f -> Function[{x, y}, x y]
Out
y
You can also try Hold and Release or Defer etc.
Hold@D[f[x, y], x] /. {f[x, y] -> x*y}
D[x y, x]
Hold@D[f[x, y], x] /. {f[x, y] -> x*y} // Release
y