Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 0 down vote accepted


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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.