Show that: $\sinh^{-1}(x) = \ln(x + \sqrt{x^2 +1 } )$ could someone Please give me some hint of how to do this question thanks

 A: HINT: $y=\cosh x=\frac12(e^x+e^{-x})$. Let $t=e^x$. Hence $t^2-2yt+1=0$ and solving this equation lead to the answer. 
A: Hint:
Assuming we already know these hyperbolic functions are invertible:
$$\sinh(\log(x+\sqrt{x^2+1})):=\frac12\left(e^{\log(x+\sqrt{x^2+1})}-e^{-\log(x+\sqrt{x^2+1})}\right)=$$
$$=\frac12\left(x+\sqrt{x^2+1}-\frac1{x+\sqrt{x^2+1}}\right)=\frac12\left(\frac{x^2+x^2+1+2x\sqrt{x^2+1}-1}{x+\sqrt{x^2+1}}\right)=$$
$$=\frac12\left(\frac{2x(x+\sqrt{x^2+1})}{x+\sqrt{x^2+1}}\right)=\frac{2x}2=x$$
A: First of all, lets look at the definition of $\sinh$. It says that
$$\sinh x = \frac{1}{2}(e^x-e^{-x})$$
But, $\sinh$ is just an function (or operation). It does something to an input ($x$) and produces an output ($\sinh(x)$), but $\sinh$ in and of itself is an operation, and not a value. Looking at this analagously, $x$ is the flour, $\sinh$ is the spaghetti machine, and $\sinh (x)$ are the noodles. of course we could put something else into the spaghetti machine in which case $y$ is the Play Doh, $\sinh$ is still the spaghetti machine, and $\sinh(y)$ could be some colorful hair to put on Mr. Potato Head.
 
So, because $\sinh$ is just a function (or operation or "a thing to do"), which does not depend on its input, I could just as easily have defined it as
$$\sinh y = \frac{1}{2}(e^y-e^{-y})$$
or
$$\sinh t = \frac{1}{2}(e^t-e^{-t})$$
or even
$$\sinh \gamma = \frac{1}{2}(e^\gamma-e^{-\gamma})$$
This approach allows me to use different variable names for the definitions
$$\sinh t = \frac{1}{2}(e^t-e^{-t})$$
and
$$\sinh^{-1}x = \ln(x+\sqrt{x^2+1})$$
So, to restate the problem we have

Find $\sinh^{-1}x$, where 
$$x=\sinh t = \frac{1}{2}(e^t-e^{-t})$$


First of all, note that
$$x=\sinh t \implies t=\sinh^{-1}x$$
So we need to solve
$$\begin{align}
\sinh t &= \frac{1}{2}(e^t-e^{-t})\\
x &= \frac{1}{2}(e^{\sinh^{-1}x}-e^{-\sinh^{-1}x})\\
2x &= e^{\sinh^{-1}x}-e^{-\sinh^{-1}x}\\
2xe^{\sinh^{-1}x} &= (e^{\sinh^{-1}x}-e^{-\sinh^{-1}x})e^{\sinh^{-1}x}\\
2xe^{\sinh^{-1}x} &= (e^{\sinh^{-1}x}-e^{-\sinh^{-1}x})e^{\sinh^{-1}x}\\
2xe^{\sinh^{-1}x} &= e^{2\sinh^{-1}x}-1\\
2xe^{\sinh^{-1}x} &= (e^{\sinh^{-1}x})^2-1\\
(e^{\sinh^{-1}x})^2-2xe^{\sinh^{-1}x}-1 &= 0 \\
\end{align}$$
You should recognize the last line as a quadratic, if not just substitute
$$u=e^{\sinh^{-1}x}$$
The last equation then becomes
$$u^2-2xu-1=0$$
solving for $u$, as per the quadratic formula we have
$$u = \frac{2x\pm\sqrt{4x^2+4}}{2}$$
$$u = x\pm\sqrt{x^2+1}$$
But
$$u=x-\sqrt{x^2+1}\le 0 \text{ and }u=e^{\sinh^{-1}x}\gt 0$$
so we must reject that solution, leaving us with
$$\begin{align}
u&=x+\sqrt{x^2+1}\\
e^{\sinh^{-1}x}&=x+\sqrt{x^2+1}\\
\ln(e^{\sinh^{-1}x})&=\ln(x+\sqrt{x^2+1})\\
\sinh^{-1}x &= \ln(x+\sqrt{x^2+1})\\
\end{align}$$
