Answer
The equation of the tangent line is $y=-1$
Work Step by Step
$y=\dfrac{x^{2}-1}{x^{2}+1}$ $,$ $(0,-1)$
Apply the quotient rule to evaluate the derivative of the given expression:
$y'=\dfrac{(x^{2}+1)(x^{2}-1)'-(x^{2}-1)(x^{2}+1)'}{(x^{2}+1)^{2}}=...$
$...=\dfrac{(x^{2}+1)(2x)-(x^{2}-1)(2x)}{(x^{2}+1)^{2}}=...$
Simplify:
$...=\dfrac{2x(x^{2}+1-x^{2}+1)}{(x^{2}+1)^{2}}=\dfrac{2x(2)}{(x^{2}+1)^{2}}=\dfrac{4x}{(x^{2}+1)^{2}}$
Substitute $x$ by $0$ in the derivative found to obtain the slope of the tangent line at the given point:
$m=\dfrac{4(0)}{(0^{2}+1)^{2}}=\dfrac{0}{1}=0$
The slope of the tangent line and a point through which it passes are known. Use the point-slope form of the equation of a line formula, which is $y-y_{1}=m(x-x_{1})$ to obtain the equation of the tangent line at the given point:
$y-(-1)=(0)(x-0)$
$y+1=0$
$y=-1$
The equation of the tangent line is $y=-1$
