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Old 2019-02-08, 16:03   #171
Dr Sardonicus
 
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Feb 2017
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Quote:
Originally Posted by jvang View Post
I'm trying to figure out the inverse of this, so that I can find the derivative of the inverse: \(y = x^3-2x^2+5x\) or something very similar. You just swap the variables and solve for y, right?
You went one step too far. Don't try to "solve for y," just swap the variables, then use implicit differentiation.

Hmm. The cubic

x^3 - 2*x^2 + 5*x - a

has discriminant -27*a^2 + 148*a - 400, which is negative for any real a. So Cardano's formulas will give the real solution to

x^3 - 2*x^2 + 5*x = a

as a sum of a rational number and two real cube roots, but it will be an algebraic mess. Differentiating would produce an even bigger mess.

Besides -- implicit differentiation works just fine, even if there is no "nice" formula for the inverse function.

Last fiddled with by Dr Sardonicus on 2019-02-08 at 16:39
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