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I have a cubic polynomial

$$f(x)=a_o x^3+a_1x^2+a_2x+a_0=(x-x_0)(x-x_1)(x-x_2)$$

but no closed form for its roots, i.e., I do not have closed form expressions for $x_1$, $x_2$ and $x_3$ in terms of the coefficients $a_0$ to $a_3$. However is there a means to derive the sensitivity of the roots with respect to the coefficients, i.e., $\frac{\partial x_1}{\partial a_1}$, etc?

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If the polynomial is $P(x) = \sum_{i=0}^3 a_i x^{3-i}$, you can get the derivative of a simple root by differentiating the implicit equation $P(x) = 0$. It gives $$x^{3-i} + P^\prime(x)\frac{\partial x}{\partial a_i} = 0$$ hence $$\frac{\partial x}{\partial a_i} = - \frac{x^{3-i}}{P^\prime(x)}$$

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It should be $$a_o x^3+a_1x^2+a_2x+a_0=a_0(x-x_0)(x-x_1)(x-x_2),$$ where $a_0\neq0.$

You can use the Cardano's formula. See here: https://en.wikipedia.org/wiki/Cubic_function#Cardano's_method

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