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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! :)