Suppose there is a function defined as $$f(x)= \frac{d}{dx}g(x)$$ similar to how $1/x$ is $ \frac{d}{dx}log(x)$ or $ \frac{1}{1+x^2}$ is $ \frac{d}{dx}\arctan(x)$ or something along those lines. Is there a general process (probably using the inverse function theorem) to find $f^{-1}(x)$?

The reason I ask is because there are more difficult functions that may end up being defined this way that don't have immediately apparent relationships. You could for instance define $e^{-x^{2}}$ as $ \frac{d}{dx}erf(x)$.

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    $\begingroup$ @nikola That's not it. Take $g(x) = x^2$, for instance. Then $f(x) = 2x$ and $f^{-1}(x) = \frac12x$. That's not an antiderivative of $g$. Also, there is no need to be condecending. This site is for asking questions. The entire point is to help people learn. No matter the age or level of the asker. $\endgroup$ – Arthur Oct 1 '18 at 6:38
  • $\begingroup$ Sorry, I am blind.... $\endgroup$ – nikola Oct 1 '18 at 6:46

I will assume everything is invertible and smooth, at least locally. The following holds for every $a$ and $b$. $$ \int_{f(a)}^{f(b)} f^{-1}(x)dx=bf(b)-af(a)-\int_a^b f(x)dx \\ =(bf(b)-g(b))-(af(a)-g(a)) $$ This can be easily seen by regarding the integration as the area of the region below the graph. You may also say that from the above equation, the antiderivative of $f^{-1}(x)$ is $$ \int f^{-1} (x)dx = xf^{-1}(x)-g\circ f^{-1}(x) $$ You can check the equality by differentiating both sides.


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