simple question - derivative - quotient rule

May 2010
13
1
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?

Will someone please help?

Ted
 

skeeter

MHF Helper
Jun 2008
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North Texas
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?

Will someone please help?

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 \(\displaystyle C = \frac{1}{B}
\)
 
May 2010
13
1
Thank you for the quick, and very helpful, reply!!