How do you differentiate arcsin(sqrt(cos^2(1/x) ) using the chain rule?

Jan 7, 2016

$\frac{1}{\sqrt{1 - \left({\cos}^{2} \left(\frac{1}{x}\right)\right)}} \cdot \frac{1}{2} \left({\cos}^{2} {\left(\frac{1}{x}\right)}^{- \frac{1}{2}}\right) \cdot 2 \cos \left(\frac{1}{x}\right) \cdot \sin \left(\frac{1}{x}\right) \cdot \frac{1}{x} ^ 2$.

Explanation:

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

I have used the following rules in conjunction to reach my final answer :

$\frac{d}{\mathrm{dx}} \left({\sin}^{- 1} u \left(x\right)\right) = \frac{1}{\sqrt{\left(1 - {u}^{2}\right)}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$.

$\frac{d}{\mathrm{dx}} {\left[u \left(x\right)\right]}^{n} = n {\left[u \left(x\right)\right]}^{n - 1} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$.

$\frac{d}{\mathrm{dx}} \cos x = \sin x$.

$\frac{d}{\mathrm{dx}} {x}^{n} = n {x}^{n - 1}$.