calculus

**tsal15** · October 30th 2008, 11:59 PM

hey y'all,

I got a question (obviously)

"By expressing $y = \sinh{(x)}$ as a quadratic in the term $e^x$ , show $\sinh^{-1}(x) = \ln{(x + \sqrt{x^2 +1})}$ ."

This is my worked solution for the quadratic part:

$y = sinh(x) = \frac{1}{2}(e^x - e^{-x})$
$y * 2e^x = {[\frac{1}{2}(e^x - e^{-x}]}*2e^x = {(e^x)}^2 - 1$
$0 = e^{2x} + 2ye^x - 1$

And now for the inverse part:

I get stuck

i know the answer is $\sinh^{-1}(x) = \ln{(x + \sqrt{x^2 +1})}$ (from the question) and i know somewhere in the calculations there is some sort of logarithm cancellation i.e. $\ln{(e^x)} = x$

Can someone help? and can you tell me if my quadratic expression is incorrect? thanks

**flyingsquirrel** · October 31st 2008, 12:39 AM

Hi,

Originally Posted by tsal15

This is my worked solution for the quadratic part:

$y = sinh(x) = \frac{1}{2}(e^x - e^{-x})$
$y * 2e^x = {[\frac{1}{2}(e^x - e^{-x}]}*2e^x = {(e^x)}^2 - 1$
$0 = e^{2x} + 2ye^x - 1$

You forgot a minus sign, the last equality should be $0 = e^{2x} {\color{blue}-} 2ye^x - 1$ . Now you can, for example, let $X=\mathrm{e}^x$ , solve for $X$ and then solve for $x$ .

**tsal15** · October 31st 2008, 01:04 AM

Originally Posted by flyingsquirrel

Hi,

You forgot a minus sign, the last equality should be $0 = e^{2x} {\color{blue}-} 2ye^x - 1$ . Now you can, for example, let $X=\mathrm{e}^x$ , solve for $X$ and then solve for $x$ .

Sorry mate, I've adjusted as you said, but i got stuck again. my answer was $y = x +1$ thats obviously a wrong answer

Can you explain it a little more? thank you

**flyingsquirrel** · October 31st 2008, 01:14 AM

Originally Posted by tsal15

Can you explain it a little more? thank you

Let $X=\exp x$ . The equation $0=\exp(2x)-2y\exp x-1$ becomes $0=X^2-2yX-1$ which is a quadratic. The discriminant is $\Delta=(-2y)^2-4\times(-1)\times 1=4y^2+4=4(y^2+1)>0$ hence the two solutions we're looking for are $X=\frac{2y\pm\sqrt{4(y^2+1)}}{2}=\frac{2y\pm\sqrt{ 4}\cdot\sqrt{y^2+1}}{2}=y\pm\sqrt{y^2+1}$ .

Can you take it from here ?

**tsal15** · October 31st 2008, 01:45 AM

Originally Posted by flyingsquirrel

Let $X=\exp x$ . The equation $0=\exp(2x)-2y\exp x-1$ becomes $0=X^2-2yX-1$ which is a quadratic. The discriminant is $\Delta=(-2y)^2-4\times(-1)\times 1=4y^2+4=4(y^2+1)>0$ hence the two solutions we're looking for are $X=\frac{2y\pm\sqrt{4(y^2+1)}}{2}=\frac{2y\pm\sqrt{ 4}\cdot\sqrt{y^2+1}}{2}=y\pm\sqrt{y^2+1}$ .

Can you take it from here ?

YES YES THANK YOU

Much appreciated

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