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December 7th, 2018, 01:38 PM  #1 
Newbie Joined: Dec 2018 From: Canada Posts: 1 Thanks: 0  Function whose derivative is inversely proportional to the function itself
Is there such a function that its derivative is one divided by the function itself? This would be analogous to how the derivative of an exponantial is directly proportional to the exponential function itself, but in this case I’m looking for it to be inversely proportional. Does such a function exist at all, and if so what is it? 
December 7th, 2018, 01:59 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,663 Thanks: 649 
$y=k\sqrt{x}$.

December 7th, 2018, 02:01 PM  #3 
Senior Member Joined: Dec 2015 From: Earth Posts: 327 Thanks: 42 
Consider equation $\displaystyle y’= \frac{C}{y}$ for Cconstant
Last edited by idontknow; December 7th, 2018 at 02:04 PM. 
December 8th, 2018, 02:10 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,663 Thanks: 649 
$y'y=k$ leads to $\frac{y^2}{2}=kx$.

December 8th, 2018, 02:31 PM  #5 
Senior Member Joined: Oct 2009 Posts: 696 Thanks: 235  
December 8th, 2018, 04:13 PM  #6 
Senior Member Joined: Aug 2012 Posts: 2,135 Thanks: 621 
I had an idea that you could proceed analogously to the way you solve $f' = f$. Assume $f$ is complex analytic, apply the theorem on termbyterm differentiation of a power series, and set the coefficients of $f$ and $f'$ equal. The result gives the wellknown power series for $e^x$ [with initial condition $f(0) = 1$]. Of course you can't just take a termbyterm multiplicative inverse, so that idea doesn't work directly. However it turns out there's a formula for the multiplicative inverse of a power series, but it's a little involved. I found two nice writeups: https://math.stackexchange.com/quest...apowerseries and http://www.math.caltech.edu/~201617...r%20series.pdf. It's a special case of Faà di Bruno's formula and it involves Bell numbers. Today I learned, right? There's an awesome amount of "elementary" math that nobody knows, let alone the higher abstract stuff. If someone is so inclined, they could set the coefficients equal and work out the general form of a complex analytic function that satisfies $f' = \frac{1}{f}$. When I was looking this up I found a lot of literature for the same question about the functional inverse rather than the multiplicative inverse. There's a formula for that too. Last edited by Maschke; December 8th, 2018 at 04:21 PM. 
December 8th, 2018, 06:55 PM  #7 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,561 Thanks: 2562 Math Focus: Mainly analysis and algebra  
December 9th, 2018, 03:39 AM  #8  
Senior Member Joined: Sep 2016 From: USA Posts: 535 Thanks: 306 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
Starting from x' = 1/x, where x is assumed to be analytic, define a new variable y = 1/x. Take its derivative by the chain rule so you get \[y' = \frac{x'}{x^2} = y^3 \] Thus, you are left to solve the 2dimensional polynomial differential equation: $x' = f(x)$ where $x \in \mathbb{R}^2$ and $f(x) = (x_2,  x_2^3)$. This specific example shows up for instance, when solving the restricted nbody problem. It works even more generally than this for sin/cos, and therefore periodic functions which can be expressed in Fourier series. In general, it is used to convert an existing $N$th dimensional vector field to a polynomial vector field in higher dimensions and it works whenever the nonpolynomial part of the vector field satisfies an algebraic differential equation. Last edited by SDK; December 9th, 2018 at 03:42 AM.  

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