I need help in proving this.
$$ \frac{d}{dx}(sinh^{-1}x)=\frac{1}{\sqrt{1+x^2}} $$
I'm supposed to use the Chain Rule.
I tried proving RHS and got stuck
Let $ y=sinh^{-1}x $ therefore $ x=sinh y $ $$ x=\frac{e^x-e^{-x}}{2} $$ $$ \frac{d}{dx}=\frac{e^x+e^{-x}}{2} $$
Then I tried LHS since chain rule looks applicable there
Let $ u=\sqrt{1+x^2} $ $$ du = \frac{x}{\sqrt{1+x^2}}dx $$ $$ (\frac{d}{du}\frac{1}{u})\frac{du}{dx} $$ $$ (-\frac{1}{u^2}) \frac{x}{\sqrt{1+x^2}}$$ $$ -\frac{x}{(1+x^2)^{\frac{3}{2}}} $$
And I'm stuck.......can this even be solved?
$$y=\sinh^{-1}x$$ $$\sinh y=\sinh \sinh^{-1}x=x$$ Differentiating both sides, $$\cosh y\ dy=\ dx$$ $$\frac{dy}{dx}=\frac{1}{\cosh y}$$
We know that $\forall \ a \in \Bbb R$$$\cosh^2a=1+\sinh^2a $$
It thus follows that
$$\frac{1}{\cosh y}=\frac{1}{\sqrt{1+\sinh^2y}}=\frac{1}{\sqrt{1+x^2}}$$