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 $(1)-(4)$ ,

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

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

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

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

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