Symbolic differentiation for very long equation

Question

Alex Lee 2022-3-18

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1675064-symbolic-differentiation-for-very-long-equation

评论： Alex Lee 2022-3-21

采纳的回答： Torsten

在 MATLAB Online 中打开

Hi All,

I would like to do partial differentition for a relative long function. The function is give as following:

ni = 1;
nt = 1.4; 
R_eff = Eff_Reflection(ni,nt); %Reff represents the fraction of photons that is internally diffusely reflected at the boundary.
syms f(ua,us,g,R,rho);
D = 1 / 3 / (ua + (1-g) * us);
z0 = 1 / (ua + (1 - g) * us);
zb = (1 + R)/(1 - R) * 2 * D;
u_eff = sqrt(3 * ua * (ua + (1 - g) * us));
r1 = sqrt(z0^2 + rho^2);
r2 = sqrt((z0 + 2*zb)^2 + rho^2);
f(ua,us,g,R,rho) = z0 * (u_eff + 1/r1) * exp(-u_eff * r1) / r1^2 + (z0 + 2*zb) * (u_eff + 1/r2) * exp(-u_eff * r2) / r2^2;
Df = diff(f,ua);

The answer for running the code is:

(exp(-3^(1/2)*(ua*(ua - us*(g - 1)))^(1/2)*((1/(ua - us*(g - 1)) - (4*(R + 1))/(3*(ua - us*(g - 1))*(R - 1)))^2 + rho^2)^(1/2))*(1/(ua - us*(g - 1)) - (4*(R + 1))/(3*(ua - us*(g - 1))*(R - 1)))*((3^(1/2)*(2*ua - us*(g - 1)))/(2*(ua*(ua - us*(g - 1)))^(1/2)) + ((1/(ua - us*(g - 1)) - (4*(R + 1))/(3*(ua - us*(g - 1))*(R - 1)))*(1/(ua - us*(g - 1))^2 - (4*(R + 1))/(3*(ua - us*(g - 1))^2*(R - 1))))/(rho^2 + (1/(ua - us*(g - 1)) - (4*R + 4)/(3*(ua - us*(g - 1))*(R - 1)))^2)^(3/2)))/((1/(ua - us*(g - 1)) - (4*(R + 1))/(3*(ua - us*(g - 1))*(R - 1)))^2 + rho^2) - (exp(-3^(1/2)*(ua*(ua - us*(g - 1)))^(1/2)*((1/(ua - us*(g - 1)) - (4*(R + 1))/(3*(ua - us*(g - 1))*(R - 1)))^2 + rho^2)^(1/2))*(1/(ua - us*(g - 1))^2 - (4*(R + 1))/(3*(ua - us*(g - 1))^2*(R - 1)))*(1/((1/(ua - us*(g - 1)) - (4*(R + 1))/(3*(ua - us*(g - 1))*(R - 1)))^2 + rho^2)^(1/2) + 3^(1/2)*(ua*(ua - us*(g - 1)))^(1/2)))/((1/(ua - us*(g - 1)) - (4*(R + 1))/(3*(ua - us*(g - 1))*(R - 1)))^2 + rho^2) + (exp(-3^(1/2)*(1/(ua - us*(g - 1))^2 + rho^2)^(1/2)*(ua*(ua - us*(g - 1)))^(1/2))*(1/((1/(ua - us*(g - 1))^2 + rho^2)^(3/2)*(ua - us*(g - 1))^3) + (3^(1/2)*(2*ua - us*(g - 1)))/(2*(ua*(ua - us*(g - 1)))^(1/2))))/((1/(ua - us*(g - 1))^2 + rho^2)*(ua - us*(g - 1))) - (exp(-3^(1/2)*(1/(ua - us*(g - 1))^2 + rho^2)^(1/2)*(ua*(ua - us*(g - 1)))^(1/2))*(1/(1/(ua - us*(g - 1))^2 + rho^2)^(1/2) + 3^(1/2)*(ua*(ua - us*(g - 1)))^(1/2)))/((1/(ua - us*(g - 1))^2 + rho^2)*(ua - us*(g - 1))^2) + (2*exp(-3^(1/2)*(1/(ua - us*(g - 1))^2 + rho^2)^(1/2)*(ua*(ua - us*(g - 1)))^(1/2))*(1/(1/(ua - us*(g - 1))^2 + rho^2)^(1/2) + 3^(1/2)*(ua*(ua - us*(g - 1)))^(1/2)))/((1/(ua - us*(g - 1))^2 + rho^2)^2*(ua - us*(g - 1))^4) - (exp(-3^(1/2)*(1/(ua - us*(g - 1))^2 + rho^2)^(1/2)*(ua*(ua - us*(g - 1)))^(1/2))*(1/(1/(ua - us*(g - 1))^2 + rho^2)^(1/2) + 3^(1/2)*(ua*(ua - us*(g - 1)))^(1/2))*((3^(1/2)*(1/(ua - us*(g - 1))^2 + rho^2)^(1/2)*(2*ua - us*(g - 1)))/(2*(ua*(ua - us*(g - 1)))^(1/2)) - (3^(1/2)*(ua*(ua - us*(g - 1)))^(1/2))/((1/(ua - us*(g - 1))^2 + rho^2)^(1/2)*(ua - us*(g - 1))^3)))/((1/(ua - us*(g - 1))^2 + rho^2)*(ua - us*(g - 1))) - (exp(-3^(1/2)*(ua*(ua - us*(g - 1)))^(1/2)*((1/(ua - us*(g - 1)) - (4*(R + 1))/(3*(ua - us*(g - 1))*(R - 1)))^2 + rho^2)^(1/2))*(1/(ua - us*(g - 1)) - (4*(R + 1))/(3*(ua - us*(g - 1))*(R - 1)))*((3^(1/2)*((1/(ua - us*(g - 1)) - (4*(R + 1))/(3*(ua - us*(g - 1))*(R - 1)))^2 + rho^2)^(1/2)*(2*ua - us*(g - 1)))/(2*(ua*(ua - us*(g - 1)))^(1/2)) - (3^(1/2)*(1/(ua - us*(g - 1)) - (4*(R + 1))/(3*(ua - us*(g - 1))*(R - 1)))*(1/(ua - us*(g - 1))^2 - (4*(R + 1))/(3*(ua - us*(g - 1))^2*(R - 1)))*(ua*(ua - us*(g - 1)))^(1/2))/(rho^2 + (1/(ua - us*(g - 1)) - (4*R + 4)/(3*(ua - us*(g - 1))*(R - 1)))^2)^(1/2))*(1/((1/(ua - us*(g - 1)) - (4*(R + 1))/(3*(ua - us*(g - 1))*(R - 1)))^2 + rho^2)^(1/2) + 3^(1/2)*(ua*(ua - us*(g - 1)))^(1/2)))/((1/(ua - us*(g - 1)) - (4*(R + 1))/(3*(ua - us*(g - 1))*(R - 1)))^2 + rho^2) + (2*exp(-3^(1/2)*(ua*(ua - us*(g - 1)))^(1/2)*((1/(ua - us*(g - 1)) - (4*(R + 1))/(3*(ua - us*(g - 1))*(R - 1)))^2 + rho^2)^(1/2))*(1/(ua - us*(g - 1)) - (4*(R + 1))/(3*(ua - us*(g - 1))*(R - 1)))^2*(1/(ua - us*(g - 1))^2 - (4*(R + 1))/(3*(ua - us*(g - 1))^2*(R - 1)))*(1/((1/(ua - us*(g - 1)) - (4*(R + 1))/(3*(ua - us*(g - 1))*(R - 1)))^2 + rho^2)^(1/2) + 3^(1/2)*(ua*(ua - us*(g - 1)))^(1/2)))/(rho^2 + (1/(ua - us*(g - 1)) - (4*R + 4)/(3*(ua - us*(g - 1))*(R - 1)))^2)^2

The answer that matlab give is very very long, it the results trustable?