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I have a cubic equation of the form $$x^3-a^2x-b^2=0.$$

It is given that all roots are real, moreover, only one root is positive and the other two are negative. Let the positive root be $\alpha>0$. Can I express $\alpha$ in terms of $a,b$? In the problem $a,b$ are some function of $t$ and I have to differentiate $\alpha$ w.r.t. $t$.

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  • $\begingroup$ To only find an expression for $\alpha'$, you can differentiate the whole equation: $3\alpha^2\alpha'-2aa'\alpha-a^2\alpha'-2bb'=0$. This allows you to express $\alpha'$ in terms of $a,b,a',b',\alpha$. $\endgroup$ – Hagen von Eitzen Jun 15 '15 at 20:56
  • $\begingroup$ en.wikipedia.org/wiki/Cubic_function#General_formula_for_roots $\endgroup$ – Wolphram jonny Jun 15 '15 at 20:58
  • $\begingroup$ @Hagen von Eitzen, How do I know that I am differentiating the positive root and not the other...? $\endgroup$ – Jack Jun 15 '15 at 21:05
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You could express $\alpha$ in terms of $a$ and $b$ using the cubic formula, but the three real roots case is very messy and would not help you.

The trick to the problem is that if $$ x^3 - [a(t)]^2 x - [b(t)]^2 = 0$$ then $$ \frac{d}{dt}\left( x^3 - [a(t)]^2 x - [b(t)]^2 \right) = 0$$ so $$ 3x^2 \frac{dx}{dt} - 2a(t)\frac{da}{dt}x - [a(t)]^2\frac{dx}{dt} - 2b(t) \frac{db}{dt} = 0 $$ So your answer will be

$$\frac{d\alpha}{dt} = 2 \frac { a(t) \frac{da}{dt} \alpha + b(t) \frac{db}{dt} }{3\alpha^2 + [a(t)]^2} $$

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  • $\begingroup$ Thanks for your help. As I said to Hagen von Eitzen, this is also true for the negative roots no..? As usual I have to set this derivative equal to zero. In my problem $a,b$ are multi variable functions and I have to minimize the positive root w.r.t. them. $\endgroup$ – Jack Jun 15 '15 at 21:25

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