I have this code, looking for the derivative of this equation
syms mp dv vc c
expr2 = (1-((c^2)/(vc^2))*(exp(dv/c)-1))/(exp(dv/c))==mp;
simplify(diff(expr2,c))
I am expecting just a simple expression, but the output contains an extra item
c^2*dv - 2*c^3*exp(dv/c) + dv*vc^2 + 2*c^3 == 0 & vc ~= 0
What does the thing after the & sign mean? Why is it here?

 採用された回答

Askic V
Askic V 2023 年 1 月 25 日

0 投票

It is probably because you he a division by vc in your expression.
.../(vc^2)...
This means simplified expression is valid only if vc is different than zero.

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