Second derivative of an autonomous function

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SUMMARY

The discussion focuses on calculating the second derivative of the function defined by the first derivative dy/dt = ry ln(K/y). The user initially derived the second derivative as d^2y/dt^2 = (ry')[ln(K/y) + 1/Kln(K/y), but received feedback indicating a missing term and an incorrect differentiation of ln(K/y). The correct second derivative is d^2y/dt^2 = (ry')[ln(K/y) - 1], highlighting the importance of accurate differentiation in calculus.

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MathewsMD
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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!
 
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MathewsMD said:
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)

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.
 

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