Answer
$y=2x-1.$
Work Step by Step
Using the quotient rule: $fâ(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$
$u(x)=|x| ; u'(x)=\dfrac{x}{|x|}$
$v(x)=\sqrt{2-x^2} ; v'(x)=-\dfrac{x}{\sqrt{2-x^2}}$
$f'(x)=\dfrac{(\dfrac{x}{|x|})(\sqrt{2-x^2})-(|x|)(-\dfrac{x}{\sqrt{2-x^2}})}{2-x^2}$ $=\dfrac{2x}{|x|\sqrt{(2-x^2)^3}}$
$f'(1)=\dfrac{2(1)}{(1)\sqrt{(2-1^2)^3}}=2.$
Equation of tangent:
$(y-y_0)=m(x-x_0)$ at point $(x_0, y_0)$ and slope $m$.
$(y-1)=2(x-1)\rightarrow y=2x-1.$
A graphing utility is used to graph the function f and the specified tangent line. The intersection point at (1, 1) is confirmed.
