# Second derivative of an autonomous function

1. Sep 22, 2014

### MathewsMD

For the derivative: dy/dt = ry ln(K/y)

I am trying to solve the second derivative. It seems like an easy solution, and I did:

d^2y/dt^2 = rln(K/y)y' + ry(y/K)

which simplifies to:

d^2y/dt^2 = (ry')[ln(K/y) + 1/Kln(K/y)

Unfortunately, the answer is d^2y/dt^2 (ry')[ln(K/y) - 1] and I don't quite see where I went wrong. Any help would be greatly appreciated!

2. Sep 23, 2014

### pasmith

You're missing a $y'$ from the last term and you haven't differentiated $\ln(K/y)$ correctly. Since $$\ln(K/y) = \ln K - \ln y$$ we have $$\frac{d}{dy} (\ln(K/y)) = \frac{d}{dy} (\ln K - \ln y) = -\frac 1y.$$