How to use implicit differentiation to prove $\sinh^{-1}(3x)$ equal to something?

Question

I understand that you let $y = \sinh^{-1}(3x)$ and thus $\sinh(y) = 3x$, but not sure where to go from there.

Answer 1

Start from basic principles. The hyperbolic sine function can be expressed as

$$\sinh y=\frac{e^y-e^{-y}}{2}\tag 1$$

We can easily invert $(1)$ to find that

$$y=\log\left(\sinh y+\sqrt{\sinh^2 y+1}\right)$$

Letting $y=\text{arsinh}(x)$ reveals

$$\text{arsinh}(x)=\log\left(x+\sqrt{x^2+1}\right)$$

Note that

$$\begin{align} \cosh (\text{arsinh}(x))&=\frac12\left(x+\sqrt{x^2+1}+\frac{1}{x+\sqrt{x^2+1}}\right)\\\\ &=\frac12\left(x+\sqrt{x^2+1}+\frac{1}{x+\sqrt{x^2+1}}\frac{x-\sqrt{x^2+1}}{x-\sqrt{x^2+1}}\right)\\\\ &=\sqrt{x^2+1} \tag 1 \end{align}$$

Therefore, using the chain rule we have for $3x=\sinh(y(x))$

$$\begin{align} \frac{d(3x)}{dx}&=3\\\\ &=\frac{d\sinh(y(x))}{dx}\\\\ &=\cosh (y(x))y'(x) \tag 2 \end{align}$$

whereupon solving $(2)$ for $y'$ and using $(1)$ yields

$$\bbox[5px,border:2px solid #C0A000]{y'(x)=\frac{3}{\sqrt{(3x)^2+1}}=\frac{1}{\sqrt{x^2+\frac19}}}$$