# Thread: simple question - derivative - quotient rule

1. ## simple question - derivative - quotient rule

I'm having problems using the quotient rule and am hoping that someone can help.

g(t) = A / 1+ BE^-t

Using the quotient rule the numerator simplifies to

0 - A(-BE^-t) = ABE^-t

I don't see how (1 + BE^-t)(A') simplifies to 0,

and how -A(1 + BE^-t)' simplifies to -A(-BE^-t).

In the latter if e is raised to the -t shouldn't that come down using the power rule and then it's e ^t-1?

Ted

2. Originally Posted by tedmetro
I'm having problems using the quotient rule and am hoping that someone can help.

g(t) = A / 1+ BE^-t

Using the quotient rule the numerator simplifies to

0 - A(-BE^-t) = ABE^-t

I don't see how (1 + BE^-t)(A') simplifies to 0,

and how -A(1 + BE^-t)' simplifies to -A(-BE^-t).

In the latter if e is raised to the -t shouldn't that come down using the power rule and then it's e ^t-1?

Ted
review your rules for derivatives ...

the derivative of a constant is 0 ... $\displaystyle A$ is a constant.

the derivative of $\displaystyle e^{-t}$ is $\displaystyle -e^{-t}$ ... the power rule does not apply to exponential functions of this type.

$\displaystyle g(t) = \frac{A}{1+Be^{-t}}$

where A and B are constants.

you do not need the quotient rule to find g'(t) ... the chain rule will suffice.

rewrite as

$\displaystyle g(t) = A(1+Be^{-t})^{-1}$

$\displaystyle g'(t) = -A(1+Be^{-t})^{-2} \cdot (-Be^{-t})$

$\displaystyle g'(t) = \frac{ABe^{-t}}{(1+Be^{-t})^2}$

$\displaystyle g'(t) = \frac{A}{1+Be^{-t}} \cdot \frac{Be^{-t}}{1+Be^{-t}}$

$\displaystyle g'(t) = g(t) \cdot \frac{1}{1 + Ce^t}$ where $\displaystyle C = \frac{1}{B}$