Answer
(a) The equation of the tangent line is: $~~~y = 2x-1$
(b) We can see a sketch of the graphs below.
Work Step by Step
(a) $y = \frac{\vert x \vert}{\sqrt{2-x^2}}$
When $x \geq 1$:
$y = \frac{x}{\sqrt{2-x^2}}$
We can find $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{\sqrt{2-x^2}-\frac{1}{2}(2-x^2)^{-1/2}(-2x)(x)}{2-x^2}$
$\frac{dy}{dx} = \frac{\frac{2-x^2}{\sqrt{2-x^2}}+\frac{x^2}{\sqrt{2-x^2}}}{2-x^2}$
$\frac{dy}{dx} = \frac{2}{(2-x^2)^{3/2}}$
When $x = 1~~~$ then $~~\frac{dy}{dx} = \frac{2}{(2-1^2)^{3/2}} = 2$
The slope of the tangent line is $m=2$
We can find the equation of the tangent line:
$y-1 = 2(x-1)$
$y = 2x-1$
The equation of the tangent line is: $~~~y = 2x-1$
(b) We can see a sketch of the graphs below.