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Lightning55
Oct 1, 2009, 03:17 PM
Okay, this question is confusing me.

Find the first derivative of

(e^x - e^-x) / (e^x + e^-x)

Should I use the quotient rule? I just need a little to start on how to solve this.
I do know that the derivative of e^x is e^x. Does that mean e^-x = y' = e^-x ?

Unknown008
Oct 2, 2009, 10:30 AM
:) I see you didn't make proper use of the brackets : { and }

This is it:

\frac{(e^x - e^{-x})}{(e^x + e^{-x})}

Ok, yes, you use the quotient rule.

No, the derivative of e^-x is :

y = e^{-x}

y' = -e^-x

In fact, you are multiplying the coefficient of the 'e' by the derivative of the power. So, if

y = e^{3x^2+2x}

then

y' = (6x+2)e^{3x^2+2x}

Lightning55
Oct 2, 2009, 11:41 AM
Oh yes, I did miss some brackets. Thanks for giving me the derivative of the negative power. It makes the question easier to solve. I may post my answer later once I get it.

Unknown008
Oct 2, 2009, 11:42 AM
Ok :)

galactus
Oct 2, 2009, 01:08 PM
If I may add something. Note that the identity

tanh(x)=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}

And the derivative of tanh(x) is sech^{2}(x)

Whose identity is sech^{2}(x)=\left(\frac{2}{e^{x}+e^{-x}}\right)^{2}=\frac{4e^{2x}}{e^{4x}+2e^{2x}+1}

You could even keep it in hyperbolic form if you wish. Same thing.

Unknown008
Oct 3, 2009, 01:34 AM
I've seen that function (tanh, cosh, sinh, etc) in my calculator, under the 'hyp' button. I never really understood what that was...

galactus
Oct 3, 2009, 05:52 AM
Do you have any calc books lying around? If so, look up hyperbolic functions. You will find the info and identities there as well as on the web somewhere.

We already know the tanh(x) identity, here are a few more:

\text{hyperbolic cotangent}=coth(x)=\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}

\text{hyperbolic secant}=sech(x)=\frac{2}{e^{x}+e^{-x}}

\text{hyperbolic cosecant}=csch(x)=\frac{2}{e^{x}-e^{-x}}

sinh(x)=\frac{e^{x}}{2}-\frac{1}{2e^{x}}

cosh(x)=\frac{e^{x}}{2}+\frac{1}{2e^{x}}

e^{x}=cosh(x)+sinh(x)

e^{-x}=cosh(x)-sinh(x)

cosh^{2}(x)-sinh^{x}(x)=1

\int_{-a}^{a}e^{tx}dx=\frac{2sinh(at)}{t}

Unknown008
Oct 3, 2009, 08:11 AM
Does that mean that hyperbolic cosecant is the same as cosecant, I mean the inverse of sine, hyperbolic secant is secant, the inverse of cosine, etc?

This one

e^x = cosh(x) + sinh(x)

resembles the identity in complex numbers:

e^{i\theta} = cos\theta + isin\theta

galactus
Oct 3, 2009, 08:39 AM
No, hyperbolic cosecant is not the same as regular cosecant.

csch(\frac{\pi}{2})=.434537....

csc(\frac{\pi}{2})=1

Unknown008
Oct 3, 2009, 08:40 AM
Ok, I'll make more research on the net about it then. Thanks! :)