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Been struggling to solve this. My derivative is $au-\ln(1+u^2)$ and I need to solve where the derivative is $= 0$. How do I solve $au-ln(1+u^2)=0$? Thank you!

user729424
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  • Certainly, $u= 0$ is a clear solution: $a(0) - \ln(1+ 0^2) = 0 -0 = 0.$ – amWhy Feb 24 '20 at 23:30
  • Yes, 1 solution I found from graphing was u = 0. Wanted to check if I am missing any other solutions. –  Feb 24 '20 at 23:32
  • Just so you don't engage in the X-Y problem, you may want to include the function from which you found the derivative, and your work, just to be sure your work is correct. – amWhy Feb 24 '20 at 23:34
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    There are other solutions that occur for some values of $a$. For example, $a=-1$ yields $1$ solution, $a=-1/2$ yields $3$ solutions, $a=0$ yields $1$ solution, and $a=1/2$ yields 3 solutions again. The roots outside of $u=0$ have no explicit formula and must be computed analytically – whpowell96 Feb 24 '20 at 23:36
  • @whpowell96 Thank you for the response. how do I compute the roots outside of u=0? –  Feb 24 '20 at 23:39
  • Depends on your computing environment but it can be done using a scientific calculator or just graphically – whpowell96 Feb 24 '20 at 23:45
  • "must be computed analytically" @whpowell96 I'd say, "must be computed numerically". One popular technique, Elliot, is called Newton's Method – you'll find it in every Calculus textbook, and all over the web. – Gerry Myerson Feb 25 '20 at 00:02
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    Sorry, that was a typo. They certainly must be computed numerically. Newton's method is good but it is hard to get all the roots with Newton. Bisection might be a good choice as it's pretty easy to find places where the function has different signs – whpowell96 Feb 25 '20 at 00:44

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