# calculus

• Oct 30th 2008, 11:59 PM
tsal15
calculus
hey y'all,

I got a question (obviously)

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

This is my worked solution for the quadratic part:

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

And now for the inverse part:

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

Can someone help? and can you tell me if my quadratic expression is incorrect? thanks :)
• Oct 31st 2008, 12:39 AM
flyingsquirrel
Hi,
Quote:

Originally Posted by tsal15
This is my worked solution for the quadratic part:

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

You forgot a minus sign, the last equality should be $\displaystyle 0 = e^{2x} {\color{blue}-} 2ye^x - 1$. Now you can, for example, let $\displaystyle X=\mathrm{e}^x$, solve for $\displaystyle X$ and then solve for $\displaystyle x$.
• Oct 31st 2008, 01:04 AM
tsal15
Quote:

Originally Posted by flyingsquirrel
Hi,

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

Sorry mate, I've adjusted as you said, but i got stuck again. my answer was $\displaystyle y = x +1$ thats obviously a wrong answer :( Can you explain it a little more? thank you
• Oct 31st 2008, 01:14 AM
flyingsquirrel
Quote:

Originally Posted by tsal15
Can you explain it a little more? thank you

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

Can you take it from here ?
• Oct 31st 2008, 01:45 AM
tsal15
Quote:

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

Can you take it from here ?

YES YES THANK YOU :) :) :) Much appreciated