Is my integration of this inverse hyperbolic function correct?

Question

I'm trying to differentiate this function

$y= \tanh^{-1}\left(\frac{x}{a}\right)$

From my attempts, I got:

$\frac{x}{a}=\tanh{y}$

$\frac{1}{a}=\text{sech}^2{y}\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{1}{a.\text{sech}^2y}$

$\frac{dy}{dx}=\frac{1}{a.\sqrt {1-\frac{x^2}{a^2}}}$

Could someone please verify whether this is correct

Answer 1

$\frac{\text{d}}{\text{d}x} \tanh^{-1} u=\frac{u'}{1-u^2}$ where u is a function of x. So your answer should be $\frac{1}{a(1-\frac{x^2}{a^2})}=\frac{a}{a^2-x^2}$