I need to find the partial derivates of
$f(a,b,c) = a*(1-e^{-br})^c$
to build a Hessian used in a Gauss Newton non linear modelling. I think the partial derivate for parameter $a$ is quite trivial, but I would like to have confirmation:
$\dfrac{\partial f}{\partial a} = (1-e^{-br})^c$
For $\dfrac{\partial f}{\partial b}$ I struggle a lot more, as $b$ is contained in the brackets.
Partial derivate $\dfrac{\partial f}{\partial c}$ might be, but here I am struggling more than for $\dfrac{\partial f}{\partial a}$:
$\dfrac{\partial f}{\partial c} = log(1-e^{-br})*a*(1-e^{-br})^c $
Thanks for confirming or giving my the correct solutions.
Let $u=1-e^{-br}$. Then $$f(u)=au^c$$.
For shorthand (later in this post) let $$f_{a}=\frac{\partial f}{\partial a}$$ Consider then \begin{align} \frac{\partial f}{\partial b} &= \frac{\partial f}{\partial u}\frac{d u}{ db} \\ &= acu^{c-1} (re^{-br}) \\ &= arce^{-br}(1-e^{-br})^{c-1} \end{align} Your expression for $f_a$ is correct, and for $f_c$ conisder now the function $f(c)=au^c$ \begin{align} \frac{\partial f}{\partial c} &= \frac{\partial }{\partial c}ae^{c \ln u}\\ &= a\ln u e^{c \ln u} \\ &= a\ln(1-e^{-br})u^c \\ &= a\ln(1-e^{-br})(1-e^{-br}) \end{align} Which agrees with your expression.