Derivative of the inverse of $y=(a+bx)e^{cx}$

Question

I need to solve for the 1st derivative of the inverse of $y=(a+bx)e^{cx}$ but my calculus is a bit rusty. I know that to get the inverse function, I would have to use the Lambert W method but I think that the 1st derivative is different. Anyway, if I understand correctly, the 1st derivative of the inverse is just 1/y'(yinv). In this case that would make my answer: $$\frac 1{be^{cy}+ce^{cy}(a+by)}$$ If someone could verify this for me or point me in the right direction I would really appreciate it.

Answer 1

If $y = f(x)$, then $dy = f'(x) dx$. If you are lucky, $f'$ is easier to invert than $f$.

Let's see.

$y=(a+bx)e^{cx}$, so $dy = (a+bx)c e^{cx} + b e^{cx} = e^{cx}( ac+bcx + b) = e^{cx}( ac+b+bcx) $ and this is of the same form, so you will have to use the good old W-function to invert $y'$.

Too bad.