How do you differentiate arcsin(csc(4x)) ) using the chain rule?

$\frac{d}{\mathrm{dx}} \left({\sin}^{-} 1 \csc \left(4 x\right)\right) = 4 \cdot \sec 4 x \cdot \sqrt{1 - {\csc}^{2} 4 x}$

Explanation:

We use the formula

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

$\frac{d}{\mathrm{dx}} \left({\sin}^{-} 1 \csc \left(4 x\right)\right) = \left(\frac{1}{\sqrt{1 - {\left(\csc 4 x\right)}^{2}}}\right) \frac{d}{\mathrm{dx}} \left(\csc 4 x\right)$

$\frac{d}{\mathrm{dx}} \left({\sin}^{-} 1 \csc \left(4 x\right)\right) = \left(\frac{1}{\sqrt{1 - {\csc}^{2} 4 x}}\right) \cdot \left(- \csc 4 x \cdot \cot 4 x\right) \cdot \frac{d}{\mathrm{dx}} \left(4 x\right)$

$\frac{d}{\mathrm{dx}} \left({\sin}^{-} 1 \csc \left(4 x\right)\right) = \left(\frac{- \csc 4 x \cdot \cot 4 x}{\sqrt{1 - {\csc}^{2} 4 x}}\right) \cdot \left(4\right)$

$\frac{d}{\mathrm{dx}} \left({\sin}^{-} 1 \csc \left(4 x\right)\right) = \left(\frac{- 4 \cdot \csc 4 x \cdot \cot 4 x}{\sqrt{1 - {\csc}^{2} 4 x}}\right) \cdot \left(\frac{\sqrt{1 - {\csc}^{2} 4 x}}{\sqrt{1 - {\csc}^{2} 4 x}}\right)$

$\frac{d}{\mathrm{dx}} \left({\sin}^{-} 1 \csc \left(4 x\right)\right) = \left(\frac{- 4 \cdot \csc 4 x \cdot \cot 4 x \cdot \sqrt{1 - {\csc}^{2} 4 x}}{- {\cot}^{2} 4 x}\right)$

$\frac{d}{\mathrm{dx}} \left({\sin}^{-} 1 \csc \left(4 x\right)\right) = 4 \cdot \sec 4 x \cdot \sqrt{1 - {\csc}^{2} 4 x}$

God bless....I hope the explanation is useful.