I have an implicit function and I would like to find the value of $h$ that maximizes $R$, i.e, I want to find $h$ that satisfies $\frac{\partial R}{\partial h} = 0$. The function is,
$C=\frac{A}{1+\alpha \times exp(-\beta[\frac{180}{\pi}\arctan(\frac{h}{R})-\alpha])}+10\times log(h^2+R^2)+B$,
where $C,A,\alpha,\beta$ and $B$ are constants.
What I have done is as following;
Let $x=\frac{h}{R}$, we can rewrite the above equation as
$C=\frac{A}{1+\alpha \times exp(-\beta[\frac{180}{\pi}\arctan(x)-\alpha])}+10\times \log(\frac{1}{R^2}) +10\times \log(1+x^2)+B$,
Now, taking the derivative with respect to $x$ yields,
$\frac{-20\times x}{\ln (10)}-\frac{180\times A\times \alpha \times \beta \times exp(-\beta[\frac{180}{\pi}\arctan(x)-\alpha]) }{\pi[1+\alpha \times exp(-\beta[\frac{180}{\pi}\arctan(x)-\alpha])]^2}=0$
Let $f(x)=\frac{-20\times x}{\ln (10)}-\frac{180\times A\times \alpha \times \beta \times exp(-\beta[\frac{180}{\pi}\arctan(x)-\alpha]) }{\pi[1+\alpha \times exp(-\beta[\frac{180}{\pi}\arctan(x)-\alpha])]^2}$
Now, we find the root of the function $f(x)=0$ numerically. This should give $x$ that satisfies $f(x)=0$. My questions are
(1) Is this approach correct?.If yes, how can we find $h$ that maximizes $R$ from $x$.
(2) Is there an easier way to find $h$?.
Thank you.