Substituting the partial differentiation

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Tae Woo Lee 2023-7-25

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https://ww2.mathworks.cn/matlabcentral/answers/2000426-substituting-the-partial-differentiation

评论： Tae Woo Lee 2023-7-25

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I am trying to substitute the partial differential equation but I can't find appropriate function.

Here is the sample code I have to modify.

syms u v x(u,v) y(u,v) z(x,y) 
z = exp(x*y) ;
x = 2*u+v ;
y = u/v ; 
dx = diff(x,u)
dy = diff(y,u)
dz_du = diff(z,u)
dz_du = subs(dz_du, {diff(x,u),diff(y,u)}, {dx, dy})

The result is

dx =
 
2
 
dy =
 
1/v
 
dz_du(u, v) =
 
exp(x(u, v)*y(u, v))*(y(u, v)*diff(x(u, v), u) + x(u, v)*diff(y(u, v), u))
 
dz_du(u, v) =
 
exp(x(u, v)*y(u, v))*(y(u, v)*diff(x(u, v), u) + x(u, v)*diff(y(u, v), u))

The point is, those diff(x(u, v), u) & diff(y(u, v), u) haven't substituted.

(what I want is the result below)

exp(x(u, v)*y(u, v))*(y(u, v)*2 + x(u, v)*1/v)

*At first, I changed the last line

subs(dz_du, {diff(x(u,v),u),diff(y(u,v),u)}, {dx, dy})

but it only returns error.

** I found some answers asking the version of my MATLAB and it is R2021a

Is there any way to guide matlab to substitute those partial differentiations?

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VBBV 2023-7-25

It seems you have defined z = exp(x*y) first. You should define it after x and y. Then do the diff(z,u) which will give the same result but in u and v. There's no need for subs in the equation