2
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Suppose:

f = a + b g[x, y]
h = D[f, {x, 2}, {y, 1}]
(*Out:= b g^(2,1)[x,y]*)
  1. In the function $h=b\frac{\partial^3g}{\partial x^2\partial y}$ I want to replace $\frac{\partial g}{\partial y}$ by $c\frac{\partial u}{\partial t}$ to obtain $h=bc\frac{\partial^3u}{\partial x^2\partial t}$.

  2. In the function $h=b\frac{\partial^3g}{\partial x^2\partial y}$ I want to replace $\frac{\partial^2 g}{\partial x^2}$ by $c\frac{\partial u}{\partial t}$ to obtain $h=bc\frac{\partial^2u}{\partial y\partial t}$.

After looking at the Fullform of partial derivatives this seems like a hopeless task. I have also read this post but it hasn't helped. Any suggestions and help is much appreciated.

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3
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Clear@Derivative
Derivative[n_, 1][g][x, y] := c D[u[x, t], {x, n}, t]
h
Clear@Derivative
Derivative[2, n_][g][x, y] := c D[u[y, t], t, {y, n}]
h

You can use Internal`InheritedBlock to avoid clearing Derivative again and again:

Clear@Derivative

Internal`InheritedBlock[{Derivative},
 Derivative[n_, 1][g][x, y] := c D[u[x, t], {x, n}, t]; h]

Internal`InheritedBlock[{Derivative},
 Derivative[2, n_][g][x, y] := c D[u[y, t], t, {y, n}]; h]
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  • $\begingroup$ +1 Nice. What does Clear@Derivative exactly do here? $\endgroup$ – Deep Oct 1 '17 at 9:38
  • $\begingroup$ @Deep Otherwise the rule we created via Derivative[2, n_][g][x, y] := c D[u[y, t], t, {y, n}] etc. will remain. Try removing Clear@Derivative and execute SubValues@Derivative and you'll know what I mean. $\endgroup$ – xzczd Oct 1 '17 at 9:44

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