The capacity of a signal is given by $C$
$$ C(S,N) = 7\ln(1+S/N) $$
Supposing $S$ and $N$ are :
$$ S(t) = 4 + \cos(4\pi t) $$
$$ N(t) = 2 + \sin(2\pi t) $$
What is the rate of change of the channel capacity, as a function of time ?
Do I need to use the rule of derivation of composite functions, which is : $$ \frac{dC}{dt} = \frac{dC}{dS} \cdot \frac{dS}{dt} +\frac{dC}{dN} \cdot \frac{dN}{dt} $$
It would give me $$ \frac{dC}{dt} = \frac{7}{S+N} \cdot(-4\pi \sin(4\pi t)) +\frac{-7S}{N(N+S)} \cdot 2\pi \cos(2\pi t) $$
or I juste have to replace S and N in the first equation like this : $$ C(t) = 7\ln\left(1+\frac{4 + \cos(4\pi t)}{2 + \sin(2\pi t)}\right) $$
and to find the rate change I could derivate this equation
Could someone tell me if I'm thinking wrong?
Thank
$$C(S(t),N(t)) = 7\ln\left(1+\frac{S}{N}\right)$$ Differentiate it with respect to $t$. Denote $\frac{d}{dt}(\dots)\to(\dots)'$ $$ C^{'}=\frac{7}{1+\frac{S}{N}}\left(\frac{S^{'}}{N}-\frac{SN^{'}}{N^2}\right)=\frac{7}{N(N+S)}(S^{'}N-N^{'}S) $$ Given that: $$ \begin{array} &S(t) = 4 + \cos(4\pi t)\\ N(t)=2+\sin(2\pi t) \end{array}\to \begin{array} &S^{'} = -4\pi\sin(4\pi t)\\ N^{'}=2\pi\cos(2\pi t) \end{array} $$ Plug it to equation: $$ C^{'}=7\frac{(-4\pi\sin(4\pi t)(2+\sin(2\pi t))-2\pi\cos(2\pi t)(4 + \cos(4\pi t)))}{(2+\sin(2\pi t))(6+\sin(2\pi t)+\cos(4\pi t))} $$ With a little bit of work you can confirm that $\frac{d}{dt}C=\frac{d}{dt}\left(7\ln\left(1+\frac{4 + \cos(4\pi t)}{2 + \sin(2\pi t)}\right)\right)$, gives the same result as above, as it theoretically should do.