Concept preview

What you will be able to do

1. 01Derive $\dfrac{d}{dx}\arcsin x=\dfrac{1}{\sqrt{1-x^2}}$ and $\dfrac{d}{dx}\arctan x=\dfrac{1}{1+x^2}$ by implicit differentiation of $\sin y=x$ and $\tan y=x$ together with a Pythagorean identity.
2. 02Recall and correctly apply all six inverse-trig derivatives, recognising that the three co-inverses ($\arccos,\arccot,\arccsc$) carry a minus sign relative to their partners.
3. 03Evaluate inverse-trig derivatives at specific inputs, describe the domains and the horizontal asymptotes of $\arctan$, and avoid mixing the radical form with the $1+x^2$ form or dropping a sign.

Start with $y=\arcsin x$, which by definition means $\sin y=x$ with the angle restricted to $y\in\left[-\tfrac{\pi}{2},\,\tfrac{\pi}{2}\right]$. Differentiate both sides of $\sin y=x$ implicitly with respect to $x$: the left side needs the chain rule, giving $\cos y\cdot\dfrac{dy}{dx}$, while the right side gives $1$. So $\cos y\cdot\dfrac{dy}{dx}=1$, which rearranges to $\dfrac{dy}{dx}=\dfrac{1}{\cos y}$. We have the derivative, but it is written in terms of the angle $y$; we want it in terms of $x$.