Answer:
![(d y)/(d x) = \frac{2}{\sqrt{(2 x+1)^(2) -1} } + ((-x)/(|x|√(1-x^2)) )) + (1) Sec h^(-1) (x)](https://img.qammunity.org/2021/formulas/mathematics/college/uxc90t0uvx3xcdjz6cp65ajyyk9egfw2pj.png)
Explanation:
Step(i):-
Given function
....(i)
we will use differentiation formulas
i) y = cos h⁻¹ (x)
Derivative of cos h⁻¹ (x)
![(d y)/(d x) = (1)/(√(x^2-1) )](https://img.qammunity.org/2021/formulas/mathematics/college/4nbk2ev9w5wfule27bfw0reo9idgzi42wi.png)
ii)
y = sec h⁻¹ (x)
Derivative of sec h⁻¹ (x)
![(d y)/(d x) = (-1)/(|x|√((x^2-1) )](https://img.qammunity.org/2021/formulas/mathematics/college/tfopbmlxcow8q5arwkle6k4paqc7ezuoq9.png)
Apply U V formula
![(d UV)/(d x) = U V^(l) + V U^(l)](https://img.qammunity.org/2021/formulas/mathematics/college/mk9rciufvltbkku0zdblkg078xx29pqqeg.png)
Step(ii):-
Differentiating equation (i) with respective to 'x'
![(d y)/(d x) = \frac{1}{\sqrt{(2 x+1)^(2) -1} } X (d)/(d x) (2 x+1) + x ((-1)/(|x|√(1-x^2)) )) + (1) Sec h^(-1) (x)](https://img.qammunity.org/2021/formulas/mathematics/college/81vu0gb7ozk2qlltmzf0o9nll4h8kzbpox.png)
![(d y)/(d x) = \frac{1}{\sqrt{(2 x+1)^(2) -1} } X (2) + ((-x)/(|x|√(1-x^2)) )) + (1) Sec h^(-1) (x)](https://img.qammunity.org/2021/formulas/mathematics/college/lg4ebol46i4ojop498j898lq0r1iqfw5pd.png)
Conclusion:-
![(d y)/(d x) = \frac{2}{\sqrt{(2 x+1)^(2) -1} } + ((-x)/(|x|√(1-x^2)) )) + (1) Sec h^(-1) (x)](https://img.qammunity.org/2021/formulas/mathematics/college/uxc90t0uvx3xcdjz6cp65ajyyk9egfw2pj.png)