Question

How do you take the derivative of this?

Answer 1

\[y=\frac{ 2 }{ (1+e^{-x}) }\]

Answer 2

My attempt:

Answer 3

\[(0-2*-e^-x) / (1+e^{-x})^2\]

Answer 4

Exponent rule might be helpful here :)\[\large\rm y=2(1+e^{-x})^{-1}\]You `could` leave it alone and do quotient rule if you like. Seems burdensome though.

Answer 5

Ah yes, that's what you did :) Quotient rule hehe

Answer 6

hehe

Answer 7

but product rule seems more efficient

Answer 8

ooohh

Answer 9

Not product, simple power rule.\[\large\rm \frac{d}{dx}2(1+e^{-x})^{-1}\quad=\quad 2\frac{d}{dx}(1+e^{-x})^{-1}\]

Answer 10

\[2ex^{-x}(1+e^-x )^{-2}\]

Answer 11

So we have chain rule,\[\large\rm 2\frac{d}{dx}(1+e^{-x})^{-1}\quad=\quad 2(-1)(1+e^{-x})^{-2}\frac{d}{dx}e^{-x}\]I think something got mixed up in yours, hmm

Answer 12

\[\large\rm 2\frac{d}{dx}(1+e^{-x})^{-1}\quad=\quad 2(-1)(1+e^{-x})^{-2}\left(-e^{-x}\right)\]

Answer 13

When you do exponents in LaTeX equation tool,
you have to put { } around the stuff when typing more than one character in the exponent.

Examples:
e^x is fine
e^2x is not.
you would need to write e^{2x}

Answer 14

I agree that the Power Rule, with Chain Rule, is the easiest tool to apply here.

Answer 15

Here, \[\large\rm 2\frac{d}{dx}(1+e^{-x})^{-1}\quad=\quad 2(-1)(1+e^{-x})^{-2}\left(-e^{-x}\right)\]
zepdrix has correctly applied the Power and Chain rules.