Question

y = a / 1 + be^-x ; (a,b constant) get dy/dx

Answer 1

y = a/(1+be^(-x)) is this what you meant?

Answer 2

\[y=\frac{a}{1+be ^{-x}}\]

Answer 3

derivative of
\[\frac{1}{f}\] is
\[\frac{-f'}{f^2}\]

Answer 4

\[f(x)=1+b^{-x}\]
\[f'(x)=-b^{-x}\ln(b)\]

Answer 5

should get
\[\frac{ab^{-x}\ln(b)}{(1+b^{-x})^2}\]

Answer 6

oh man was completely wrong. i thought it said
\[b^{-x}\] but it says
\[be^{-x}\] let me start again

Answer 7

\[f(x)=1+be^{-x}\]
\[f'(x)=-be^{-x}\]
derivative is
\[\frac{abe^{-x}}{(1+be^{-x})^2}\]