Q.Using the inverse-function rule, show that $\dfrac{d}{dx}\left(\sin^{-1} x\right) = \dfrac{1}{\sqrt{1 - x^2}}$, and hence find its value at $x = \tfrac12$.

Set $y=\sin^{-1}x$ so $x=\sin y$, differentiate w.r.t. $y$, then reciprocate.

Let $y = \sin^{-1} x$. By definition this means $x = \sin y$ (with $y$ in the principal range $[-\tfrac{\pi}{2}, \tfrac{\pi}{2}]$, where $\cos y \ge 0$).

Differentiate $x = \sin y$ with respect to $y$.

Express $\cos y$ in terms of $x$. Since $\sin y = x$ and $\cos y \ge 0$ on the principal range, $\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$.

Apply the inverse-function rule $\dfrac{dy}{dx} = \dfrac{1}{\,dx/dy\,}$:

Evaluate at $x = \tfrac12$.

$\frac{dy}{dx}\Big|_{x=1/2} = \frac{1}{\sqrt{1 - \tfrac14}} = \frac{1}{\sqrt{\tfrac34}} = \frac{1}{\tfrac{\sqrt3}{2}} = \frac{2}{\sqrt3}.$ …

Once derived, the result $\tfrac{d}{dx}\sin^{-1}x=\tfrac{1}{\sqrt{1-x^2}}$ is a standard result (§4) and can be quoted directly without re …

A frequent slip is taking $\cos y = \pm\sqrt{1-x^2}$ and keeping the negative sign; on the principal range of $\sin^{-1}$ the cosine is non-negative, so the positive root is correct. Another is fo …