Timeline for FindRoot of Derivative
Current License: CC BY-SA 4.0
7 events
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| Mar 3, 2020 at 14:45 | comment | added | Michael E2 |
@AlexeiBoulbitch Note that FindRoot[x^2 - 2, {x, 1}] works without an explicit Equal. (See the first paradigm in the docs for FindRoot.)
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| Mar 3, 2020 at 14:41 | comment | added | Michael E2 |
D[f[x, y1, y2], y1] won't work if y1 is a number. Search the site; there are many solutions presented. Depending on f, fd[x_?NumericQ, y1_?NumericQ, y2_?NumericQ] = D[f[x, y1, y2], y1] might work. So might fd[x_?NumericQ, y1_?NumericQ, y2_?NumericQ] := Derivative[0,1,0][f][x, y1, y2], but it's probably better to compute the derivative once and use the result (again, depends on f).
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| Mar 3, 2020 at 14:26 | comment | added | user2799609 |
Thanks, I added that fd should be equal to 0. And yes, x will be assigned a number to evaluate fr.
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| Mar 3, 2020 at 14:24 | history | edited | user2799609 | CC BY-SA 4.0 |
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| Mar 3, 2020 at 14:16 | comment | added | Alexei Boulbitch |
I think you are missing two things. 1. That FindRoot works with equations. So, you have to add Equal to your code. 2. That FindRoot is a completely numerical procedure. Therefore, all your parameters must be assigned to some number. Are you going to assign x to something?
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| Mar 3, 2020 at 14:14 | history | edited | user2799609 | CC BY-SA 4.0 |
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| Mar 3, 2020 at 14:03 | history | asked | user2799609 | CC BY-SA 4.0 |