How do you differentiate # g(x) = sqrt(arcsec(x+1) #?

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Answer 1 · 2016-07-25T13:28:51

#1/[4|x+1|sqrt(x(x+2)arcsec(x+1))]#.

Let, #y=g(x)=sqrtu, where, u=arcsect, &, t=x+1#.

Thus, #y# is a function of #u#, #u# of #t#, and #t# of #x#.

By the Chain Rule, we have,

#dy/dx=dy/(du)*(du)/(dt)*dt/dx...............(1)#

Now, #y=sqrtu rArr dy/(du)=1/(2sqrtu).....(2)#.

#u=arcsect rArr (du)/dt=1/{|t|sqrt(t^2-1)}................(3)#

#t=x+1 rArr dt/dx=1............(4)#

Therefore, by #(1)-(4)#,

#dy/dx=1/(2sqrtu)*1/{|t|sqrt(t^2-1)}*1#

#={1/(2sqrt(arcsect))}*1/[|x+1|sqrt{(x+1)^2-1}]#

#=1/(2sqrt(arcsec(x+1))){1/(2|x+1|sqrt(x^2+2x))}#

#=1/[4|x+1|sqrt(x(x+2)arcsec(x+1))]#.