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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?

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2 Answers 2

up vote 0 down vote accepted

Yes.

The outer function is $s\mapsto \displaystyle\frac1s$ or you can call its variable anything. And '$\cdot -bce^{-bt}$' should be in parenthesis: $\cdot (-bce^{-bt})$, else it seems correct.

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Beside to @Berci's answer, the following approach may be easier. It is based on chain rule as well: $$ y = \frac{\frac{b}{a}}{1+ce^{-bt}}=\frac{b}{a}(1+ce^{-bt})^{-1}\to\\\ y'=\frac{b}{a}(-1)(c)(-b)(e^{-bt})(1+ce^{-bt})^{-2}= \frac{\frac{b^2}{a}ce^{-bt}}{(1+ce^{-bt})^2}$$ as you achieved before.

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