I was told to solve for $x$ in $g(x) = x^3+2x-10$ using the cubic equation here ().
So I took $a = 1, b = 0, c = 2, d = -10$, plugged into the cubic formula and got $x =$$(5+\sqrt{\frac{683}{27}})^{\frac{1}{3}} + (5-\sqrt{\frac{683}{27}})^{\frac{1}{3}}$. Then I take the derivative, $g'(x) = 3x^2 +2$ and plug what I have from x up above into g'(x). That gets me $3((5+\sqrt{\frac{683}{27}})^{\frac{2}{3}}+(5-\sqrt{\frac{683}{27}})^{\frac{2}{3}}+2(25-\frac{683}{27})^{\frac{1}{3}})+2$. That's about $11.8$ in WolframAlpha. Then, I took the inverse derivative rule, which was finding $\frac {d} {dx} [f^{-1}(8)]$. I got $\frac{1}{14}$, which is nowhere close to $11.8$. I could really use some help.
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$\begingroup$What you wanted was a function that gives $x$ in terms of $g(x)$. So you should have taken $d = -10 - g$ in your cubic formula.
This gives three awful-looking roots, one of which is real (it is easy to tell that the equation has only one real root, because the first derivative of $g(x)$ is never zero so the function is always increasing).
$$ x(g) = \frac{\left(90+9g+\sqrt{81g^2+1620g+8196}\right)^{1/3}}{\sqrt[3]{18}} - \frac{\sqrt{16/3}}{\left(90+9g+\sqrt{81g^2+1620g+8196}\right)^{1/3}} $$
Now you should proceed by taking the derivative of that messy root with respect to $g$, finding the reciprocal of that expression, plugging in $g=x^3+2x-10$, and simplifying, hopfully getting $\frac{dg}{dx} = 3x^2+2$.
Even full Mathematica does not perform this simplification automagically; you have to guide and coax it!
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