Find dx/dy in terms of y when y(x)=(e^x)-(e^-x)

I know dy/dx=(e^x)+(e^-x)

I'm trying to get x in terms of y.

Question

Find dx/dy in terms of y when y(x)=(e^x)-(e^-x)

I know dy/dx=(e^x)+(e^-x)

I'm trying to get x in terms of y.

loser66 · Answer

he asks for $\dfrac{dx}{dy}$

loser66 · Answer

to me, I convert $sinh x = \dfrac{e^x-e^{-x}}{2}$ $\rightarrow $ 2 sinh x = $e^x -e^{-x}$ =y

loser66 · Answer

from then, I have y/2 = sinh x $\rightarrow$  x = 1/2 arcsinh (y)
then, take derivative , you have dx/dy = $\dfrac{1}{2(\sqrt{1-y^2)}}$

anonymous · Answer

May I ask, where did you get sinhx from?

loser66 · Answer

that is the form of it, kind of formula. That question is equivalent to "why 1+1=2"

loser66 · Answer

look at this http://tutorial.math.lamar.edu/Classes/CalcI/DiffHyperTrigFcns.aspx

anonymous · Answer

I see, so I have to dx/dy : 1/2 arcsinh y.

loser66 · Answer

you have formula for this, too. the problem is just convert, and that's all.

anonymous · Answer

Can you show me how this is done? or the relevant formula. Thanks.

loser66 · Answer

look at this http://math.info/Calculus/Derivatives_Hyp_InvHyp/

anonymous · Answer

The answer is $$\frac{ 1 }{ \sqrt{y ^{2}+4} }$$

It is also possible to do this by taking $$y ^{2} and (\frac{ dy }{ dy })^2 $$ and comparing the two results.

loser66 · Answer

don't get what you mean