y = arcsinhx = Ln(x+sqrt((x^2) + 1)))

[gadav478]

Good evening-

My professor proved that y = arcsinhx = Ln(x+sqrt((x^2) + 1)))

I think I'm missing an algebraic step.

He has that x = sinhy = ((e^y) - (e^-y))/2
I get that...

Then he substitutes e^y for t.

So he has x = ((t^2) -1)/(2t).
I get that...

He then uses the quadratic formula to find what t equals but I am not sure how he got ((t^2) -1)/(2t) into a quadratic.

He then has: t = (2x +- sqrt(4x^2 +4))/4 for the quadratic formula. This is where I'm confused.

After that, I understand the simplification and substitution but the introduction of the quadratic formula is confusing. Can anyone see how he did that?

Thanks for your help.

 

[gadav478]

Good evening-

My professor proved that y = arcsinhx = Ln(x+sqrt((x^2) + 1)))

I think I'm missing an algebraic step.

He has that x = sinhy = ((e^y) - (e^-y))/2
I get that...

Then he substitutes e^y for t.

So he has x = ((t^2) -1)/(2t).
I get that...

He then uses the quadratic formula to find what t equals but I am not sure how he got ((t^2) -1)/(2t) into a quadratic.

He then has: t = (2x +- sqrt(4x^2 +4))/4 for the quadratic formula. This is where I'm confused.

After that, I understand the simplification and substitution but the introduction of the quadratic formula is confusing. Can anyone see how he did that?

Thanks for your help.

I understand it now. It looks like he used "2x" as a variable in the quadratic as well. That is what confused me. He simply multiplied both sides by 2x.

For the quadratic: a = 1, b= -2x and c = -1

 

[DrPhil]

I understand it now. It looks like he used "2x" as a variable in the quadratic as well. That is what confused me. He simply multiplied both sides by 2x.

For the quadratic: a = 1, b= -2x and c = -1

Actually multiplied both sides by 2t, giving (2x)t on the left.