Suppose we have a strictly increasing and continuous function $f(x)$ such that for some $\bar{x}: \exists\ f'(\bar{x}) = 0$ while the derivative exists and is positive in a small neighborhood of $\bar{x}$. Function $y(x)$ is given implicitly by $$af(x)+bf(y)=(a+b)f(\bar{x})$$ where $a, b$ are some positive constants. I am interested in $\frac{dy(x)}{dx} \bigg|_{x = \bar{x}}$.
From the Implicit function theorem we know that $$\frac{dy(x)}{dx} = -\frac{af'(x)}{bf'(y)}$$ as long as $f'(y) \neq 0$ which is not true at the point $x = \bar{x}$, since $y(\bar{x}) = \bar{x}$ and $f'(\bar{x}) =0$.
My question is: Suppose that the derivative above actually exists, moreover there exist finite, non-zero left and right second derivatives $f''_{+}(\bar{x})$ and $f''_{-}(\bar{x})$ which have different signs.
Would it be true to write: $$\frac{dy(x)}{dx}\biggl|_{x=\bar{x}+0} = \lim_{x \to \bar{x}+0} -\frac{af'(x)}{bf'(y(x))}\ ?$$ And then to use L'Hospital rule to get $$\frac{dy(x)}{dx}\biggl|_{x=\bar{x}+0} = \lim_{x \to \bar{x}+0} -\frac{af''_{+}(x)}{bf''_{-}(y(x)) y'(x)}$$ Concluding that $$\frac{dy(x)}{dx}\biggl|_{x=\bar{x}+0} = -\sqrt{\frac{a}{b}\bigg|\frac{f''_{+}(\bar{x})}{f''_{-}(\bar{x})}\bigg|}$$ where the minus sign follows from the fact that clearly $y(x)$ is decreasing since $f(x)$ is increasing and $a, b > 0$
If the conclusion above is incorrect which additional assumptions I need for it to work?
Would be very grateful for any help. Thanks in advance!
This is not a case for using l'Hospital. Instead use Taylor expansion on the two sides. Set $\lambda_+=f''_+(\bar{x})>0$ and $\lambda_-= f''_-(\bar{x})<0$.
For $h>0$: $f(\bar{x}+h) -f(\bar{x})= \frac12 \lambda_+ h^2+o(h^2) = \frac12(\lambda_+ + \epsilon(h)) h^2$.
For $k<0$: $f(\bar{x}+k) - f(\bar{x})= \frac12 (\lambda_-+\epsilon(k))k^2$.
Suppose now that $x=\bar{x}+h$ with $h>0$ (small) which implies that $y=y(x)=\bar{x}+k$ with $k=k(h)<0$. Then $$ a (\lambda_+ +\epsilon(h)) h^2 + b (\lambda_- + \epsilon(k)) k^2 = 0,$$ which implies $y'_+(\bar{x}) = \lim_{h \rightarrow 0^+} k(h)/h = -\sqrt{-\frac{a\lambda_+}{b\lambda_-}}$ similarly $y'_-(\bar{x}) = \lim_{h \rightarrow 0^-} k(h)/h = -\sqrt{-\frac{a\lambda_-}{b\lambda_+}}$. Note that the derivatives from the left and from the right may be different.