What is the derivative of $[e^x / (1 - e^x)]$ ?

Question

1841 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-07-31T11:26:42

$y^' = e^x/(1- e^x)^2$

The quotient rule will work just fine for this function because you can write it as

$y = e^x/(1-e^x) = f(x)/g(x)$

In such cases, the derivative of the function can be found by

$color(blue)(d/dx(y) = (f^'(x) * g(x) - f(x) * g^'(x))/[g(x)]^2$ , with $g(x)!=0$ .

Apart from this, all you really need to know is that

$d/dx(e^x) = e^x$

So, the derivative of $y$ will be

$d/dx(y) = ([d/dx(e^x)] * (1 - e^x) - e^x * d/dx(1 - e^x))/(1 - e^x)^2$

$y^' = (e^x * (1-e^x) - e^x * (-e^x))/(1-e^x)^2$

$y^' = (e^x - color(red)(cancel(color(black)(e^(2x)))) + color(red)(cancel(color(black)(e^(2x)))))/(1 - e^x)^2$

$y^' = color(green)(e^x/(1- e^x)^2)$

What is the derivative of [e^x / (1 - e^x)][e^x / (1 - e^x)]?