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

Jul 25, 2016

$\frac{1}{4 | x + 1 | \sqrt{x \left(x + 2\right) a r c \sec \left(x + 1\right)}}$.

#### Explanation:

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,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}} \ldots \ldots \ldots \ldots \ldots \left(1\right)$

Now, $y = \sqrt{u} \Rightarrow \frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{2 \sqrt{u}} \ldots . . \left(2\right)$.

$u = a r c \sec t \Rightarrow \frac{\mathrm{du}}{\mathrm{dt}} = \frac{1}{| t | \sqrt{{t}^{2} - 1}} \ldots \ldots \ldots \ldots \ldots . \left(3\right)$

$t = x + 1 \Rightarrow \frac{\mathrm{dt}}{\mathrm{dx}} = 1. \ldots \ldots \ldots . . \left(4\right)$

Therefore, by $\left(1\right) - \left(4\right)$,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{u}} \cdot \frac{1}{| t | \sqrt{{t}^{2} - 1}} \cdot 1$

$= \left\{\frac{1}{2 \sqrt{a r c \sec t}}\right\} \cdot \frac{1}{| x + 1 | \sqrt{{\left(x + 1\right)}^{2} - 1}}$

$= \frac{1}{2 \sqrt{a r c \sec \left(x + 1\right)}} \left\{\frac{1}{2 | x + 1 | \sqrt{{x}^{2} + 2 x}}\right\}$

$= \frac{1}{4 | x + 1 | \sqrt{x \left(x + 2\right) a r c \sec \left(x + 1\right)}}$.