What is the derivative of f(x)=arcsin sqrt sinx?

Jul 3, 2015

One can derive the derivative for $\arcsin x$ with implicit differentiation if it is not easy to remember it.

$y = \arcsin x$
$\sin y = x$
$\cos y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 1$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\cos y} = \frac{1}{\sqrt{1 - {\sin}^{2} y}} = \frac{1}{\sqrt{1 - {x}^{2}}}$

since ${\sin}^{2} x + {\cos}^{2} x = 1$.

Thus, take this further with the Chain Rule.

$\frac{d}{\mathrm{dx}} \left[\arcsin \sqrt{\sin x}\right] = \frac{1}{\sqrt{1 - {\left(\sqrt{\sin} x\right)}^{2}}} \cdot \frac{1}{\left(2 \sqrt{\sin} x\right)} \cdot \cos x$

$= \textcolor{b l u e}{\cos \frac{x}{2 \sqrt{\sin} x \sqrt{1 - \sin x}}}$