Question #1cabe

Apr 21, 2017

$f ' \left(x\right) = 4 \arcsin \left(x\right) \frac{1}{\sqrt{1 - {x}^{2}}}$

Explanation:

$f ' \left(x\right) = 4 \arcsin \left(x\right) \left(\arcsin \left(x\right)\right) '$

To calculate $\left(\arcsin \left(x\right)\right) '$ use the inverse rule:

$y = {f}^{- 1} \left(x\right) \setminus R i g h t a r r o w y ' = \frac{1}{f ' \left(y\right)}$

Hence if $y = \arcsin \left(x\right)$, we have $x = \sin \left(y\right) , x ' = \cos \left(y\right)$, hence $y ' = \frac{1}{\cos} \left(y\right) = \frac{1}{\cos} \left(\arcsin \left(x\right)\right)$.

In order to compute $\cos \left(\arcsin \left(x\right)\right)$, consider

${\cos}^{2} \left(\arcsin \left(x\right)\right) + {\sin}^{2} \left(\arcsin \left(x\right)\right) = 1$, but ${\sin}^{2} \left(\arcsin \left(x\right)\right) = {x}^{2}$, hence

$\cos \left(\arcsin \left(x\right)\right) = \sqrt{1 - {x}^{2}}$. Hence we get the answer