I want to find $y'$ where $ y = \frac{\frac{b}{a}}{1+ce^{-bt}}.$ But I dont want to use quotient rule for differentiation. I want to use chain rule. My solution is: Write $y=\frac{b}{a}\cdot \frac{1}{1+ce^{-bt}}.$ Then in $\frac{1}{1+ce^{-bt}},$ the inner function is $1+ce^{-bt}$ and the outer function is $\frac{1}{1+ce^{-bt}}.$ Hence using the chain rule we have $\left(\frac{1}{1+ce^{-bt}}\right)'= \frac{-1}{(1+ce^{-bt})^2} \cdot -bce^{-bt} = \frac{bce^{-bt}}{(1+ce^{-bt})^2}.$
Thus $y'= \frac{\frac{b^2}{a}ce^{-bt}}{(1+ce^{-bt})^2}.$ Am I correct?