Composite & Quotient Rule -confirm correct answer?

looking0glass · #1 $Old$ June 9th, 2009, 03:39 AM

Hi,

I would be really grateful if someone would help me!

With regard to the function f(x) = ln(e^x + e^-x)

By applying the composite rule does this turn into:

1 / e^x+ e^-x ?

And by applying the quotient rule, does the first equation turn into:

e^x + e^ -x / e^x - e^ -x

Many thanks!!!

pickslides · #2 $Old$ June 9th, 2009, 04:43 PM

You don't need to use the quoient rule in this case, consider

$\frac{d}{dx}(ln(f(x)) = \frac{f'(x)}{f(x)}$

In your case $f(x) = e^x + e^{-x}$ and

$f'(x) = e^x - e^{-x}$ therefore

$\frac{d}{dx}(ln(e^x + e^{-x})) = \frac{e^x - e^{-x}}{e^x + e^{-x}}$

HallsofIvy · #3 $Old$ June 9th, 2009, 05:23 PM

Quote:

Originally Posted by looking0glass $View Post$

Hi,

I would be really grateful if someone would help me!

With regard to the function f(x) = ln(e^x + e^-x)

By applying the composite rule does this turn into:

1 / e^x+ e^-x ?

And by applying the quotient rule, does the first equation turn into:

e^x + e^ -x / e^x - e^ -x

Many thanks!!!

Please don't just say "turn into"- I would interpret that as meaning a different form of the same equation or function!

The composite rule (also called "chain rule") tells us that the derivative of f(x)= [math]ln(e^x+ e^{-x})[/itex] is $\frac{1}{e^x+ e^{-x}}$ times the derivative of $e^x+ e^{-x}$ which is $e^x- e^{-x}$ .

So the derivative of $f(x)= ln(e^x+ e^{-x})$ is $f'(x)= \frac{e^x- e^{-x}}{e^x+ e^{-x}}$ , the reciprocal of what you gave.

That derivative is also, by the way, $tanh(x)$ .