Inverse Function
- (mathematics) For a given function ƒ, another function, denoted \(f^{-1}\), that reverses the mapping action of ƒ; (formally) given a function , a function such that,
Synonyms
- (function that reverses the mapping action of a given function): anti-function (obsolete or nonstandard in this sense)
\[y=f(x)\]
\[x=f^{-1}(y)\]
\[1=\frac {df^{-1}(y)}{dy}\frac {dy}{dx}\]
\[\frac {df^{-1}(y)}{dy}=\frac {dx}{dy}\]
\[\frac {df^{-1}(y)}{dy}=\frac {dx}{dy}\]
\[\frac {df^{-1}(y)}{dy}=\frac1\frac {dy}{dx}\]
\[y=sin (x)\]
\[x=\arcsin (y)\]
\[1=\frac {d \arcsin (y)}{dy}.\cos (x)\]
\[\frac {d \arcsin (y)}{dy}=\frac1{\cos (x)}\]
\[\frac {d \arcsin (y)}{dy}=\frac1{\sqrt{1-\sin^2 (x)}}\]
\[\frac {d \arcsin (y)}{dy}=\frac1{\sqrt{1-y^2}}\]
\[y=\cos (x)\]
\[x=\arccos (y)\]
\[1=-\frac {d \arccos(y)}{dy}.\sin (x)\]
\[\frac {d \arccos (y)}{dy}=-\frac1{\sin (x)}\]
\[\frac {d \arcsin (y)}{dy}=-\frac1{\sqrt{1-\cos^2 (x)}}\]
\[\frac {d \arccos (y)}{dy}=-\frac1{\sqrt{1-y^2}}\]
\[y=\tan (x)\]
\[x=\arctan (y)\]
\[1=\frac {d \arctan (y)}{dy}.[1+\tan^2 (x)]\]
\[\frac {d \arctan (y)}{dy}=\frac1{1+\tan^2 (x)}\]
\[\frac {d \arcsin (y)}{dy}=\frac1{1-y^2}\]
\[y=\log_a x\]
\[x=a^y\]
\[1=\]