Editing Generalised logistic function (section)

== Gradient of generalized logistic function ==
When estimating parameters from data, it is often necessary to compute the partial derivatives of the logistic function with respect to parameters at a given data point <math>t</math> (see<ref name=fekedulegn1999parameter>{{cite journal|last=Fekedulegn|first=Desta|author2=Mairitin P. Mac Siurtain|author3=Jim J. Colbert|title=Parameter Estimation of Nonlinear Growth Models in Forestry|journal=Silva Fennica|year=1999|volume=33|issue=4|pages=327–336|doi=10.14214/sf.653|url=http://www.metla.fi/silvafennica/full/sf33/sf334327.pdf|access-date=2011-05-31|archive-url=https://web.archive.org/web/20110929005929/http://www.metla.fi/silvafennica/full/sf33/sf334327.pdf|archive-date=2011-09-29|url-status=dead}}</ref>).  For the case where <math>C = 1</math>, 
:<math>
\begin{align}
\\
\frac{\partial Y}{\partial A} &= 1 - (1 + Qe^{-B(t-M)})^{-1/\nu}\\
\\
\frac{\partial Y}{\partial K} &= (1 + Qe^{-B(t-M)})^{-1/\nu}\\
\\
\frac{\partial Y}{\partial B} &= \frac{(K-A)(t-M)Qe^{-B(t-M)}}{\nu(1 + Qe^{-B(t-M)})^{\frac{1}{\nu}+1}}\\
\\
\frac{\partial Y}{\partial \nu} &= \frac{(K-A)\ln(1 + Qe^{-B(t-M)})}{\nu^2(1 + Qe^{-B(t-M)})^{\frac{1}{\nu}}}\\
\\
\frac{\partial Y}{\partial Q} &= -\frac{(K-A)e^{-B(t-M)}}{\nu(1 + Qe^{-B(t-M)})^{\frac{1}{\nu}+1}}\\
\\
\frac{\partial Y}{\partial M} &= -\frac{(K-A)QBe^{-B(t-M)}}{\nu(1 + Qe^{-B(t-M)})^{\frac{1}{\nu}+1}}
\\
\end{align}
</math><!-- DY/dt missing .... -->