I'm not quite sure how to phrase this, but does anyone know what the property (see image attached) is called?
$$\left.\frac{df^{-1}}{dx}\right|_{f(x)}=\frac{1}{df/dx|_a}$$
There may be a mistake somewhere but I didn't even know how to begin searching for it on Google without a name. Thanks!
If $y=f(x)$, is one-to-one and onto (bijection) in domain $D$ so $x=f^{-1}(y)$ exists and let $(x_0,y_0)$ be a good point on the curve such that $y_0=f(x_0)$. Then $$\left .\frac{df^{-1}(x)}{dx}\right|_{x=y_0}=\left .\frac{df^{-1}(y)}{dy}\right|_{y=y_0}=\frac{1}{f'(x_0)}.$$
Suppose $y=f(x)=x^5+x+1,$ it is bijection (1,3) and (2,35) are good points whereas $y=1,2,4,5,6,7,8,...,32,34,..$ cannot give a good point as we cannot find $x_0$ forthem by HAND.
$$\left .\frac{df^{-1}(x)}{dx}\right|_{x=3}=\frac{1}{6}, \left .\frac{df^{-1}(x)}{dx}\right|_{x=33}=\frac{1}{81}.$$
But we cannot find $$\left .\frac{df^{-1}(x)}{dx}\right|_{x=1,2,4,5.6,7,8,......32,34,...}= ?$$ cannot be found by HAND.
This interesting property does not have any name and it can be proved by noting that $$f^{-1}(f(x))=x.$$