$y=\sinh ^{-1}\left(\frac{x}{a}\right)$ I am struggling with two things regarding the derivative of sinh, but also other functions like it. 
$$y=\sinh ^{-1}\left(\frac{x}{a}\right)$$
$$x=a \sinh (y)$$
Know my question is, if take the derivative regarding x I get:
$$1=a \cosh (y)*\frac{\text{dy}}{\text{dx}}$$
But do I really?
$$\frac{d\sinh ^{-1}\left(\frac{x}{a}\right)}{dx}$$
=$$\frac{\frac{1}{a}-\frac{x \frac{da}{dx}}{a^2}}{\sqrt{\frac{x^2}{a^2}+1}}$$
My first question is therefore is it the derivative or the partial derivative I am taking? Second question I understand that I can use pythagorean theorem such that $$\cosh (y)=\sqrt{\sinh ^2(y)+1}$$ However, how can I substitute $$\sinh ^2(y)=x^2$$
Please reply to both questions
 A: It's definitely the ordinary derivative because you are differentiating with respect to one independent variable. 
And for the result : 
$$y=\sinh ^{-1}\left(\frac{x}{a}\right)\\\implies\frac{dy}{dx}=\dfrac{1}{a\times \sqrt{\left(\frac{x}{a}\right)^2+1}}\\\implies\frac{dy}{dx}=\dfrac{1}{\sqrt{x^2+a^2}}.$$  
Also see the list of derivatives of inverse hyperbolic functions here.
A: I think you are expected to treat $a$ a constant so $\frac{da}{dx}=0$ and the partial derivative is the same as the derivative.
From $\frac{x}{a}=\sinh(y)$ and $1=a \cosh (y)\frac{\text{d}y}{\text{d}x}$ you get 
$$\frac{\text{d}y}{\text{d}x}=\frac{1}{a \cosh (y)} = \frac{1}{a \sqrt{\sinh ^2(y)+1}}= \frac{1}{a \sqrt{\frac{x^2}{a^2}+1}}$$ which is the same as your other approach, assuming $a$ is constant.
A: You can rewrite $$y=\sinh ^{-1}\left(\frac{x}{a}\right)$$ as $$x=a \sinh(y)$$ $$\frac{dx}{dy}=a \cosh(y)$$ $$\frac{dy}{dx}=\frac{1}{a \cosh(y)}$$ and since $\cosh(y)=\sqrt{1+\sinh^2(y)}=\sqrt{1+\frac{x^2}{a^2}}$, then $\cdots$
I am sure that you can take from here.
