I am trying to find the derivative with respect to T of: $$\frac FV(T_{\mathrm{in}} - T)+\beta k_0 \exp\left(-\frac{EaR}{T}\right)\left(C_{A,\mathrm{in}}+\frac{1}{\beta}\left(T_{\mathrm{in}}-T\right)\right)\left(C_{B,\mathrm{in}}+\frac{2}{\beta}\left(T_{\mathrm{in}}-T\right)\right).$$
This is what I came up with: $$-\frac FV + \frac{\beta k_0 EaR \exp\left(-\frac{EaR}{T}\right)C_{A,\mathrm{in}}\cdot C_{B,\mathrm{in}}}{T^2} - 2\left(C_{A,\mathrm{in}}+\frac{1}{\beta}\left(T_{\mathrm{in}}-T\right)\right) - \left(C_{B,\mathrm{in}}+\frac{2}{\beta}\left(T_{\mathrm{in}}-T\right)\right).$$
Obviously the derivative of $\frac FV \left(T_{\mathrm{in}} - T\right)$ is $-\frac FV$.
And the derivative of the terms $C_{A,\mathrm{in}}+\frac{1}{\beta}\left(T_{\mathrm{in}}-T\right)$ and $C_{B,\mathrm{in}}+\frac{2}{\beta}\left(T_{\mathrm{in}}-T\right)$ are $-\frac{1}{\beta}$ and $-\frac{2}{\beta}$ respectively.
The derivative of $\beta k_0 \exp\left(-\frac{EaR}{T}\right)$ is
$$\frac{\beta k_0 (-EaR)}{T^2}.$$
From there I simply use the chain rule to gather it all, but I just can't figure whether I got it right. Can anyone give me a second opinion on the derivative?
Thanks
We should have
$$-(F/V)+\\ +(E_aR/T^2)\cdot\beta\cdot k_0\cdot \exp(-E_aR/T)\cdot\left(C_{A,in}+\frac{1}{\beta}\cdot(T_{in}-T)\right)\cdot\left(C_{B,in}+\frac{2}{\beta}\cdot(T_{in}-T)\right)+\\ +\beta\cdot k_0\cdot \exp(-E_aR/T)\cdot\left(-\frac{1}{\beta}\right)\cdot\left(C_{B,in}+\frac{2}{\beta}\cdot(T_{in}-T)\right)+\\ +\beta\cdot k_0\cdot \exp(-E_aR/T)\cdot\left(C_{A,in}+\frac{1}{\beta}\cdot(T_{in}-T)\right)\cdot\left(-\frac{2}{\beta}\right)$$
which can be reordered as
$$-(F/V)+\beta\cdot k_0\cdot \exp(-E_aR/T)\cdot\\\left[(E_aR/T^2)\cdot\left(C_{A,in}+\frac{1}{\beta}\cdot(T_{in}-T)\right)\cdot\left(C_{B,in}+\frac{2}{\beta}\cdot(T_{in}-T)\right)+ \left(-\frac{1}{\beta}\right)\cdot\left(C_{B,in}+\frac{2}{\beta}\cdot(T_{in}-T)\right)+ \left(C_{A,in}+\frac{1}{\beta}\cdot(T_{in}-T)\right)\cdot\left(-\frac{2}{\beta}\right)\right]$$
Note that the expression is in the form
$$f(T)=a(T)+b(T)\cdot c(T)\cdot d(T)$$
therefore
$$f'(T)=a'(T)+b'(T)\cdot c(T)\cdot d(T)+b(T)\cdot c'(T)\cdot d(T)+b(T)\cdot c(T)\cdot d'(T)$$